## (1)

I've just realized that I did not read the question carefully, so an edit is in order.

## (a) $\mathscr G$ is flat over $S$

So, let's see the real question: Let $\mathscr F\to \mathscr G$ be a morphism of coherent sheaves on $A$ with kernel $\mathscr K$ and cokernel $\mathscr C$. In other words one has an exact sequence
$$
0\to \mathscr K \to \mathscr F \to \mathscr G \to \mathscr C \to 0.
$$
Base changing to $k(s)$ gives an exact sequence
$$
\mathscr F_s \to \mathscr G_s \to \mathscr C_s \to 0.
$$
If $\mathscr F_s \to \mathscr G_s$ is an isomorphism, then $\mathscr C_s=0$ and if this holds for all $s\in S$, then $\mathscr C=0$. (Nakayama's lemma implies that if $\mathscr C_s=0$, then $\mathscr C$ is zero along $f^{-1}(s)$, hence $\mathrm{supp} \mathscr C\subset A\setminus f^{-1}(s)$. If this holds for all $s\in S$, then $\mathrm{supp}\mathscr C=\emptyset$, so $\mathscr C=0$).

So now we have a short exact sequence
$$
0\to \mathscr K \to \mathscr F \to \mathscr G \to 0.
$$
Base changing to $k(s)$ gives the exact sequence
$$
\mathscr K_s \to \mathscr F_s \to \mathscr G_s \to 0,
$$
and if $\mathscr G$ is flat over $S$, then this last exact sequence is also exact on the left. In that case, again, if $\mathscr F_s \to \mathscr G_s$ is an isomorphism, then $\mathscr K_s=0$ and if this holds for all $s\in S$, then $\mathscr K=0$ (same way as above).

## (b) $\mathscr G$ is not flat over $S$

Here is a simple example to show that $\mathscr G$ being flat (or something else perhaps) is a necessary condition. Let $S$ be arbitrary, $E$ an abelian variety and $A=E\times S$. Let $p\in E$, $t\in S$, and $Z=\{p\}\times S$. Further let $\mathfrak m\subseteq \mathscr O_Z$ be the maximal ideal corresponding to the point $(p,t)\in Z$. Now let $\mathscr F=\mathscr O_Z/\mathfrak m^2$, $\mathscr G=\mathscr O_Z/\mathfrak m$, and $\mathscr F \to \mathscr G$ the natural projection. Then $\mathscr F_s \to \mathscr G_s$ is an isomorphism for every $s\in S$: Indeed if $s\neq t$, then $\mathscr F_s = \mathscr G_s=0$ and for $s=t$, they are both isomorphic to $\mathscr O_Z/\mathfrak m$ and the restriction of the above morphism *is* an isomorphism, but the original $\mathscr F \to \mathscr G$ is not injective.

## (c) other stuff

Originally I did not read the question carefully and thought you just want to obtain information about $\mathscr F$ and $\mathscr G$ from how $\mathscr F_s$ and $\mathscr G_s$ behave. In other words, if there weren't a morphism $\mathscr F \to \mathscr G$ given, then the outlook is much more bleak. Actually, in part (2) it is not clear whether you have such a morphism or not, so this might still be relevant for that case.

If you do not have a morphism $\mathscr F \to \mathscr G$ given, you would always have to account for twisting one of the sheaves by a line bundle pulled back from $S$: For any $s\in S$, $\mathscr F_s\simeq (\mathscr F\otimes f^*\mathscr N)_s$ where $f:A\to S$ is the structure map and $\mathscr N$ is a line bundle on $S$.

There is something along these lines that is actually true:

Assume that $S$ is integral of finite type over an algebraically closed field, $f$ is flat and projective and the fibers are integral. Let $\mathscr L$ and $\mathscr M$ be two line bundles on $A$ such that $\mathscr L_s\simeq \mathscr M_s$ for all $s\in S$. Then there exists a line bundle $\mathscr N$ on $S$ such that $\mathscr L\simeq \mathscr M\otimes \mathscr N$.

To prove this, apply cohomology and base change to the sheaf $f_*(\mathscr L\otimes \mathscr M^{-1})$. I suppose you can try to generalize the method of proving this to a little bit more general situation, but you probably need more information to get much further.

## (2)

Let $Z=e(S)$. Then $f|_Z:Z\to S$ is an isomorphism and in particular $\mathscr O_Z$ is flat over $S$. So, if you have a morphism $\mathscr F\to \mathscr O_Z$, then you can apply (1). If you don't have such a morphism, then there is the problem I mentioned above.