The ring $\hat{\mathbb{Z}}$ can be defined in several ways, including as the product of the $p$-adic integers, $\mathbb{Z}_p$, over all primes $p$, or as the inverse limit of the rings $\mathbb{Z}/n\mathbb{Z}$ with morphisms between rings $\mathbb{Z}/n\mathbb{Z}\rightarrow \mathbb{Z}/d\mathbb{Z}$ when $d\mid n$.
This latter definition can be rephrased as follows. Let $R$ be the ring of functions $f: \mathbb{N}^+\rightarrow \mathbb{Z}$ such that $\forall n,m\in \mathbb{N}^+$, $f(n)\equiv f(m) \pmod{\operatorname{gcd}(n,m)}$.
Let $I=\left\{f\in R: \forall n\in\mathbb N^+,n\mid f(n)\right\}$. $I$ is an ideal of $R$.
Then $R/I \cong \hat{\mathbb{Z}}$.
Now, clearly, $R\supset \operatorname{Cong}(\mathbb{N}^+)$, and it turns out that $\operatorname{Cong}(\mathbb{N}^+)/(I\cap\operatorname{Cong}(\mathbb{N}^+)) \cong \hat{\mathbb{Z}}$.
In general, if $S\subset \mathbb{Z}$ and $n\in \mathbb{Z}$, we can define an ideal in $\operatorname{Cong}(S)$: $$\mathcal{J}_{S,n} = \{f\in \operatorname{Cong}(S): \forall s\in S, (s-n)\mid f(s)\}$$ and define the quotient ring: $$O_{S,n} = \operatorname{Cong}(S)/{\mathcal{J}_{S,n}}$$
We will call these rings "obstruction rings," because they measure the difficulty of extending a function on $S$ to a value at $n$.
When $S$ is infinite, the ring always contains $\mathbb{Z}$ in a trivial way. Let $k\in \mathbb{Z}$, then define $c_k(s)=k$ to be the constant function on $S$. Then $c_k\in\operatorname{Cong}(S)$, and $c_j-c_k\in\mathcal{J}_{S,n}$ if and only if $j=k$.
An element $f\in\operatorname{Cong}(S)$ that maps to an image of an integer, $k$, can be extended to $f(n)=k$, so we can think of the obstruction ring as the ring of "possible values at $n$ of congruent functions on $S$."
If $n\in S$, then $O_{S,n} \cong \mathbb{Z}$, with the isomorphism defined by sending $f\in\operatorname{Cong}(S)$ to $f(n)$.
If $n\notin S$, $\mathcal{J}_{S,n}$ is the set of $f\in\operatorname{Cong}(S)$ which can be extended to $n$ by defining $f(n)=0$, and $O_{S,n}$ is isomorphic to some infinite image of $\hat{\mathbb{Z}}$
The $O_{S,n}$ can be represented as rings of the form: $$\prod_p{R_p}$$ where the $R_p$ is either all of the $p$-adic integers, $\mathbb{Z}_p$, or it is $\mathbb{Z}/p^k\mathbb{Z}$ for some $k\geq 0$. (So, in particular, $R_p=\{0\}$ is allowed.)
In order for this ring to be infinite, either some $R_p$ is all of $\mathbb{Z}_p$, or there are infinitely many $p$ such that $R_p\neq \{0\}$.
We can find $R_p$ based on $S$ and $n\notin S$ by taking $R_p=\mathbb{Z}/p^k\mathbb{Z}$ if $p^k$ is the highest power of $p$ which divides some $s-n$, or $R_p=\mathbb{Z}_p$ if no such highest power exists.
In particular, since a non-trivial product ring is never an integral domain, $\mathcal{J}_{S,n}$ is prime only when exactly one of $R_p=\mathbb{Z}_p$ and all the rest are $\{0\}$. For that to be the case, $S-n$ must only consist of powers of $p$.
We can also see that $O_{S,n}$ is uncountable if any only if $R_p=\mathbb{Z}_p$ for some $p$, which is equivalent to showing that $n$ is is the $p$-adic closure of $S$ for some $p$.
We could also think of $O_{S,n}$ as an inverse limit of the rings $\mathbb{Z}/d\mathbb{Z}$ restricted to $\{d: \exists s\in S: d\mid (s-n)\}$.
If $S\subset T$, the natural map $\operatorname{Cong}(T)\rightarrow\operatorname{Cong}(S)$ determines a homomorphism $O_{T,n}\rightarrow O_{S,n}$, which is onto if $n\notin T$.
Theorem: $O_{\mathbb{N}^+,0} \cong \hat{\mathbb{Z}}$
Proof: If $f\in\operatorname{Cong}(\mathbb{N}^+)$, for all $d>0$ we can define $f_d\in{\mathbb{Z}/d\mathbb{Z}}$ as $f_d= f(d)+d\mathbb{Z}$. Since $d\mid f(d)-f(dk)$ , these are compatible with the inverse limit definition, so $\{f_d\}$ corresponds to an element of $\hat{\mathbb{Z}}$. $f$ is in the kernel of this homomorphism if and only if $f(d)=f_d\in d\mathbb{Z}$ for all $d$, which is equivalent to $f\in \mathcal{J}_{\mathbb{N}^+,0}$, so this induces a 1-1 homomorphism from $O_{\mathbb{N}^+,0}$ to $\hat{\mathbb{Z}}$.
Now to prove this map is onto. An element of $\hat{\mathbb{Z}}$ is uniquely determined by a set of values $\{a_{p^k}\in \mathbb{Z}/p^k\mathbb{Z}\}$ with the condition that $a_{p^{k+1}} \equiv a_{p^k} \pmod {p^k}$. Given such a set, we'll define $f\in\operatorname{Cong}(\mathbb{N}^+)$ which has the property that $f(p^k)\equiv a_{p^k} \pmod {p^k}$.
We will use our classification of the elements of $\operatorname{Cong}(\mathbb{N})$ as being in the form: $$f(n) = \sum_{i=0}^{n} b_i \frac{\operatorname{lcm}[i]}{i!} (n)_i$$ Where $\operatorname{lcm}[m]=\operatorname{lcm} \{1,2,..,m\}$. So an element of $g\in \operatorname{Cong}(\mathbb{N}^+)$ can be written as: $$g(n) = \sum_{i=1}^{n} b_i \frac{\operatorname{lcm}[i-1]}{(i-1)!} (n-1)_{i-1}$$
Let $u_i(k) = \frac{\operatorname{lcm}[i-1]}{(i-1)!} (k-1)_{i-1}$. Then $u_i\in\operatorname{Cong}(\mathbb{N}^+)$, and $u_i(k)=0$ when $k<i$, and $u_i(i) = \operatorname{lcm}[i-1]$.
We will define $b_i$ inductively so that $g(p^k)\equiv a_{p^k} \pmod {p^k}$ for all $p^k$.
Define $b_i=0$ if $i$ is not a positive power of a prime.
Assume $b_1,...,b_{p^k-1}$ are defined.
Let $g_{p^k}(n) = \sum_{i=1}^{p^k-1} b_iu_i(n)$
Then $g_{p^k}$ is congruent, so $g_{p^k}(p^k)\equiv g_{p^k}(p^{k-1}) = g_{p^{k-1}}(p^{k-1}) \equiv a_{p^{k-1}} \equiv a_{p^k} \pmod {p^{k-1}}$
So we want to define $b_{p^k}$ so that: $$g_{p^k}(p^k) + b_{p^k} u_{p^k}(p^k) \equiv a_{p^k} \pmod {p^k}$$
But $u_{p^k}(p^k)=\operatorname{lcm}[p^k-1]$, so this is equivalent to finding $b_k$ so that: $$g_{p^k}(p^k) + b_{p^k} \operatorname{lcm}[p^k-1] \equiv a_{p^k} \pmod {p^k}$$
Now, subtract $g_{p^k}(p^k)$ from both sides: $$b_{p^k} {\operatorname{lcm}[p^k-1]} \equiv a_{p^k}-g_{p^k}(p^k) \pmod {p^k}$$
But we saw that $a_{p^k} - g_{p^k}(p^k)$ is divisible by $p^{k-1}$ and so is $\operatorname{lcm}[p^k-1]$, so dividing on both sides, we get: $$b_{p^k} {\frac{\operatorname{lcm}[p^k-1]}{p^{k-1}}}\equiv A \pmod p$$ for a specific integer A.
But now we can solve for $b_{p^k}$ because we know that $\frac{\operatorname{lcm}[p^k-1]}{p^{k-1}}$ is not divisible by $p$. We can, in fact, choose $b_{p^k} \in \{0,1,...,p-1\}$
Corollary: $O_{\mathbb{Z}-\{0\},0} \cong \hat{\mathbb{Z}}$
Proof: If $f\in \operatorname{Cong}(\mathbb{N}^+)$, define $f'\in \operatorname{Cong}(\mathbb{Z}-\{0\})$ such that $f'(n)=f(n^2)$. If $f'$ in $\mathcal{J}_{\mathbb{Z}-\{0\},0}$, then $n\mid f'(n)=f(n^2)$ for all $n\neq 0$. But $f(n^2)\equiv f(n) \pmod n$ for all $n>0$, so $f\in\mathcal{J}_{\mathbb{N}^+,0}$. On the other hand, if $f\in\mathcal{J}_{\mathbb{N}^+,0}$, $f'\in \mathcal{J}_{\mathbb{Z}-\{0\},0}$
Therefore, we get an induced map $\hat{\mathbb{Z}} \cong O_{\mathbb{N}^+,0} \rightarrow O_{\mathbb{Z}-\{0\},0}$ which is necessarily 1-1.
Now, we need to show that this map is onto.
Let $h\in\operatorname{Cong}(\mathbb{Z}-\{0\})$. Let $f\in\operatorname{Cong}(\mathbb{N}^+)$ be just the restriction of $h$ to $\mathbb{N}^+$. Then $f'(n)=h(n^2)$, and we want to show that $f'-h \in \mathcal{J}_{\mathbb{Z}-\{0\},0}$. But $f'(n)-h(n) = h(n^2)-h(n) \equiv 0 \pmod n$, so we are done.
Lemma: If $S$ is infinite, and $0\notin S$, then the inclusion $\operatorname{Cong}(\mathbb{Z}-\{0\})\rightarrow\operatorname{Cong}(S)$ induces a homomorphism $\hat{\mathbb{Z}} \cong O_{\mathbb{Z}-\{0\},0}\rightarrow O_{S,0}$ which is onto.
Proof: If $f\in\mathcal{J}_{\mathbb{Z}-\{0\},0}$, then clearly the restriction of $f$ to $S$ is in $\mathcal{J}_{S,0}$, so the induced map exists.
To prove that this map is onto, we need to take an arbitrary $g\in\operatorname{Cong}(S)$, and find an $f\in\operatorname{Cong}(\mathbb{Z}-\{0\})$ such that $s\mid f(s)-g(s)$ for all $s\in S$.
Let $g_{p^k}\in \mathbb{Z}/p^k\mathbb{Z}$ be defined for prime powers such that $p^k\mid s$ for some $s\in S$ as $g(s)+p^k\mathbb{Z}$. This definition is well-defined because $g(s)\equiv g(s') \pmod {p^k}$ when $s\equiv s' \pmod {p^k}$. These $g_{p^k}$ satisfy the condition that $g_{p^{k-1}}\equiv g_{p^k} \pmod {p^{k-1}}$. We can extend the set of $g_{p^k}$ to all prime powers easily enough. As we saw before, there is a member $f\in\operatorname{Cong}(\mathbb{Z}-\{0\})$ such that $f(p^k)\equiv g_{p^k} \pmod {p^k}$.
Now, if $s\in S$, we know that $g(s)\equiv g_{p^k} \equiv f(p^k) \equiv f(s) \pmod {p^k}$ if $p^k\mid s$. By the Chinese Remainder Theorem, this means that $g(s)\equiv f(s) \pmod s$, and so $f$ satisfies our condition.
Theorem: If $S\subset T$, and $n\notin T$, then there is an induced map $O_{T,n}\rightarrow O_{S,n}$ which is onto.
Proof: The proof that there is an induced map comes from the same reasoning as above.
That it is onto is due to the fact that the map from $\operatorname{Cong}(\mathbb{Z}-\{n\})$ to $O_{S,n}$ factors through $O_{T,n}$.
Let $S$ be infinite, $m,n\notin S$. When can we find an $f\in\operatorname{Cong}(S)$ which can be extended to $m$ and to $n$, but not compatibly?
If, for every $p^\alpha\mid (m-n)$, there is an $s\in S$ such that $p^\alpha\mid (s-n)$, then we can see that $m$ and $n$ are always compatible. This condition is equivalEnt to $O_{S\cup\{m\},n} \cong O_{S,n}$.
This condition is also necessary. Since $O_{S\cup\{m\},n}\rightarrow O_{S,n}$ is onto, if it is not an isomorphism, then there is an $f$ in $\operatorname{Cong}(S\cup\{m\})$ which is not in $\mathcal{J}_{S\cup\{m\},n}$ but whose image in $\operatorname{Cong}(S)$ is in $\mathcal{J}_{S,n}$. But that means that $f_{\mid S}$ can be extended to $S\cup\{n\}$ with $f(n)=0$, but not to $S\cup\{m,n\}$.
This also shows that if $O_{S\cup\{m\},n} \cong O_{S,n}$, then $O_{S\cup\{n\},m} \cong O_{S,m}$
In a sense, these theorems say that the obstructions to extending a congruent function are as big as you might expect. Given $f\in\operatorname{Cong}(S)$, and $0\notin S$, there is only one possible value for $f(0)$ modulus $p^k$ when $p^k\mid s$, for some $s\in S$. Our theorems show that given a compatible set of values $a_{p^k}\in \mathbb{Z}/p^k\mathbb{Z}$, we can always find an $f$ for which $f(0)$ must be $a_{p^k}$ mod $p^k$. The image of $f$ in $O_{S,0}$ can be thought of as the "value" of $f$ extended to zero, with this extension represents a true congruent function if and only if the image of $f$ is in the natural embedding of $\mathbb{Z}$ in $O_{S,0}$.
Similarly, two values $m,n\notin S$ are compatible only if they are "obviously" compatible - if the prime powers that divide $m-n$ do not add any prime power additions to the prime powers dividing some $s-n$. For example, if $m,n$ are even, and all the values of $S$ are odd, then there is an $f\in\operatorname{Cong}(S)$ which can be extended to $m$ and to $n$, but incompatibly.
If $S=\{k^2: k\in \mathbb{Z}\}$ is the set of perfect squares, and $n$ is even and square-free, then: $$O_{S,n} = \mathbb{Z}/n\mathbb{Z} \times \prod_{(n/p)=1} \mathbb{Z}_p$$
If $n$ is odd and square-free, it's a bit more complicated - the question of $p=2$ is a bit harder. And for $n$ which has square factors, it gets even more complicated.
If $P$ is the set of prime numbers, and $n$ not a prime: $$O_{P,n} = \left(\prod_{p\mid n} \mathbb{Z}/p\mathbb{Z}\right) \times \left(\prod_{p{\not\mid } n} \mathbb{Z}_p\right)$$
This equality is a result of Dirichlet's prime number theorem.
Note that, if $n$ is square-free, we get a formula not unlike the formula for squares when $n$ is square-free: $$O_{P,n} = \mathbb{Z}/n\mathbb{Z} \times (\prod_{p{\not\mid } n} \mathbb{Z}_p)$$
Notice that $O_{P,n}$ is countable if and only if $n=0$. Is it possible for an infinite set $S$ to have two values $n$ such that $O_{S,n}$ is countable? In fact, we can construct such a set where all of the obstruction rings are countable. Let $$S=\{0\}\cup \{m!: m\in \mathbb{N}\}$$
Since the $p$-adic limit of $m!$ is $0$ for all primes $p$, this means that for any $n\notin S$, and any $p$, $n$ is not in the $p$-adic closure of $S$. But that means that there must be a maximum value of $k$ such that there exists an $s\in S$ such that $p^k\mid s-n$. So in this case, $O_{S,n}$ is countable for all $n$.
$O_{S,n}$ is uncountable if and only if $n\notin S$ and $n$ is in the $p$-adic closure of $S$ for some $p$.
When $S$ was infinite, we saw that the for $f\in \operatorname{Cong}(S)$, the image of $f$ in $O_{S,n}$ could be thought of as the "value of $f$ at $n$."
We can still take that view when $S$ is finite, because the possible values for $f$ at $n$ are actually a congruence class, modulo $D$.
In general, if $n\notin S$, we have a map $\phi_{S,n}:\hat{\mathbb{Z}}\rightarrow {O}_{S,n}$ which is onto. The kernel of this map, which we might call $\hat{\mathcal{J}}_{S,n}$, can be written as: $$\bigcap_{s\in S} (s-n)\hat{\mathbb{Z}}$$
But the intersection of ideals is the natural generalization of the least common multiple. It's just that for infinite $S$, the ideal: $$\bigcap_{s\in S} (s-n)\mathbb{Z}$$ is always trivial, so we have to expand the ring $\mathbb{Z}$ so that we can get non-trivial least common multiples of infinite sets.
Also, if $D\in\mathbb{Z}-\{0\}$, then $\hat{\mathbb{Z}}/D\hat{\mathbb{Z}} \cong \mathbb{Z}/D\mathbb{Z}$. So $O_{S,n}$ is an image of $\hat{\mathbb{Z}}$ whenever $n\notin S$.