First let's spell out what it means to preserve the equivalence relation. I assume that it means that if $x\sim y$, then $f(x)\sim f(y)$.
Monomorphisms
Note that monomorphisms are preserved by right adjoint functors. Proof of the dual statement at stacks, currently Proposition 4.5.
Let $U:\newcommand\MSet{\mathbf{MSet}}\MSet \to \newcommand\Set{\mathbf{Set}}\Set$ be the forgetful functor, $U(S,\sim_S)=S$.
I claim that the functor $L:\Set\to \MSet$ defined by $LS = (S,=_S)$ is left-adjoint to $U$.
This is more or less immediate, since if $(T,\sim_T)\in \MSet$ (which by abuse of notation I will also call $T$), then any function $f:S\to T$ respects the equivalence relation on $LS$, so we have $$\newcommand\Hom{\operatorname{Hom}}\Hom_\Set(S,UT)=\Hom_\Set(S,T)\simeq\Hom_\MSet(LS,T).$$
Thus $U$ preserves monomorphisms. Hence if $f$ is a monomorphism in $\MSet$, it must be a monomorphism in $\Set$. Therefore all monomorphisms are injective.
It should also be fairly clear that if $f$ is an injective function that respects the equivalence relation, then it is going to be a monomorphism for the same reason that it is a monomorphism in $\Set$.
Epimorphisms
Similar to monomorphisms, surjective functions that respect the equivalence relation are epimorphisms in $\MSet$. To show that these are all of the epimorphisms it would be good if the forgetful functor had a right adjoint as well, and indeed it does.
Let $RS = (S,S\times S)$. Then just like with the left adjoint, all functions from a set with an equivalence relation to $RS$ respect the equivalence relations, so $$\Hom_\Set(US,T)\simeq \Hom_\MSet(S,RT).$$
Edit
The proofs given above can be rephrased in elementary language, and it's frankly a bit cleaner (probably because I thought these proofs out thoroughly before writing them this time).
The elementary proof:
Proof.
First, as above, injective and surjective functions that respect the equivalence relation are still monomorphisms and epimorphisms respectively, so we just need to prove that monomorphisms (epimorphisms) are injective (surjective).
Let $f:(S,\sim_S)\to (T,\sim_T)$ be a monomorphism. We'll prove that $f$ is a monomorphism in $\Set$. Suppose that we have a set $A$ and arrows $a,b : A\to S$ such that $f\circ a = f\circ b$. Then $a$ and $b$ induce morphisms $a,b:(A,=_A)\to (S,\sim_S)$. Thus $f\circ a=f\circ b$ as morphisms $(A,=_A)\to (T,\sim_T)$ in $\MSet$. Then since $f$ is a monomorphism in $\MSet$, $a=b$ as morphisms in $\MSet$, which means that $a=b$ as functions. Thus $f$ is a monomorphism in $\Set$.
The proof when $f$ is an epimorphism is symmetric, but we now have parallel functions $a,b : T\to A$, and they induce morphisms $(T,\sim_T)\to (A,A\times A)$ this time. $\quad\blacksquare$
There's a standard trick using the Yoneda lemma for computing what universal objects in (restricted) functor categories must be if they exist. In the case of the subobject classifier, this is explained on p. 37 of Sheaves in Geometry and Logic.
Specifically, suppose that $\Omega\colon \mathbf{N}\to \mathbf{FinSets}$ is a subobject classifier in $\mathbf{FinSets}^\mathbf{N}$. Let's try to find out what finite set $\Omega(0)$ is. Since $\mathbf{FinSets}^\mathbf{N}$ is a full subcategory of $\mathbf{Sets}^\mathbf{N}$, by Yoneda we have $$\Omega(0) \cong \text{Hom}_{\mathbf{Sets}^\mathbf{N}}(h^0,\Omega) = \text{Hom}_{\mathbf{FinSets}^\mathbf{N}}(h^0,\Omega) \cong \text{Sub}(h^0),$$ where $h^0$ is the functor $\text{Hom}_{\mathbf{N}}(0,-)$. But now $h^0$ has infinitely many subobjects, but $\Omega(0)$ is a finite set, and this is a contradiction.
To see that $h^0$ has infinitely many subobjects, just note that since $0$ is the initial object in $\mathbf{N}$, $h^0(n)$ is a singleton $\{*\}$ for all $n$. Incidentally, this makes $h^0$ isomorphic to the terminal object $1$ - your guess about the identity of terminal object is correct.
Now for each natural number $n$ (or $n = \infty$), there is a distinct subobject of $h^0$, given by $$m\mapsto \begin{cases} \varnothing & \text{if }m<n\\ \{*\} & \text{if }m\geq n.\end{cases}$$
Best Answer
You aren't trying to prove anything for all objects $X.$ You're trying to prove that $\phi(a_1)\ne \phi(a_2)$ when $a_1\ne a_2,$ using the assumption that for all objects $X$ and all functions $x,y:X\to A,$ $\phi\circ x=\phi\circ y\implies x=y.$ In particular, this assumption holds when $X=\{e\},$ which is all that is needed for the result. The converse of your theorem is thus the harder direction.