If $f$ is congruent on an infinite set $S_1$, and $g$ is congruent on an infinite set $S_2$, and $S=S_1\cap S_2$ is infinite and $f=g$ on $S$, then can we extende $f$ to $S_0\cup S_1$?
The answer is "no." Define an $S$ as follows: $$s_0 = 3$$ $$s_{k+1} = s_k(s_k-1)(s_k-2)+1$$
Lemma: The set of values $\{s_k(s_k-2)\}$ are pairwise relatively prime.
Proof: If $p$ divides $s_i(s_i-2)$, it is easy to show by induction that for $k>i$ that $s_k \equiv 1 \pmod p$, and hence that $p$ does not divide $s_k(s_k-2)$.
Theorem: If $S=\{s_0,...,s_k,...\}$ as defined here, we can define $f\in\operatorname{Cong}(S)$ such that $f(s_k)\equiv 0 \pmod {s_k}$ and $f(s_k)\equiv 1 \pmod {s_k-2}$ for all $k$. Thus $f$ can then be extended as $f(0)=0$ on $S\cup\{0\}$ and as $f(2)=1$ on $S\cup\{2\}$, but these extensions are not compatible.
Proof: Define $a_0=0.$
Given $a_0,...,a_{k-1}$, we want to define $a_k$ as follows:
Let $m = \sum_{i=0}^{k-1} a_i q_{S,i}(s_k)$
Then $f(s_k) = m + a_k q_{S,k}(s_k)$
We want to find $a_k$ so that $f(s_k)\equiv0 \pmod {s_k}$ and $f(s_k) \equiv 1 \pmod {s_k-2}$. That's equivalent to wanting to finding an $a_k$ so that: $$f(s_k) = m + a_k q_{S,k}(s_k) \equiv \frac {s_k(s_k-1)}{2} \pmod {s_k(s_k-2)}$$ (Since $s_k$ is odd, $s_k$ and $s_k-2$ are relatively prime.)
We can always find such an $a_k$ if we know that $q_{S,k}(s_k)$ is relatively prime to $s_k(s_k-2)$.
But $$q_{S,k}(s_k) = (s_k-s_0)(s_k-s_1)...(s_k-s_{k-1}) \equiv (-1)^k s_0s_1...s_{k-1} \pmod {s_k}$$
By our lemma, $s_k$ is relatively prime to $s_i$ for $i\neq k$, so $q_{S,k}(s_k)$ is relatively prime to $s_k$.
Similarly, $$q_{S,k}(s_k) \equiv (-1)^k (s_0-2)(s_1-2)...(s_{k-1}-2) \pmod {s_k-2}$$ and, again by the lemma, $s_k-2$ is relatively prime to $s_i-2$ for $i\neq k$. So $q_{S,k}(s_k)$ is relatively prime to $s_k-2$.
So, $q_{S,k}(s_k)$ is relatively prime to $s_k(s_k-2)$, and we can solve for $a_k$.
Notice, we haven't really used the full condition of the lemma, that $\{s_k(s_k-2)\}$ are pairwise relatively prime. We really only needed $\{s_k\}$ to be pairwise relatively prime, and $\{s_k-2\}$ pairwise relatively prime. If the twin prime conjecture is true, we use the top element of each pair to be $s_k$ to get a (presumably) fairly dense set $S$ for which we can find incompatible extensions at $0$ and $2$.