Since the OP suggested it, I am posting my comments as an answer. By the construction of Hilbert and Quot schemes, there is a relative Hom scheme, $\text{Hom}_S(\mathcal{A},\mathcal{B})$ over $S$ whose connected components are quasi-projective over $S$. The claim is that these components are proper over $S$. By the valuative criterion of properness, it suffices to prove this in the special case that $S$ equals $\text{Spec}(R)$ for $R$ a DVR. Denote by $K$ the fraction field of $R$. Then it suffices to prove that every $K$-morphism of $K$-fibers, $\phi_K:\mathcal{A}_K\to \mathcal{B}_K$ extends to an $R$-morphism $\phi:\mathcal{A}\to \mathcal{B}$.

This follows by the extension result of Weil. Here is the basic argument: denote by $\mathcal{A}'$ the closure in $\mathcal{A}\times_S \mathcal{B}$ of the graph of $\phi$. The projection morphism $\text{pr}_{\mathcal{A}}:\mathcal{A}'\to \mathcal{A}$ is projective and birational. Since $\mathcal{A}$ is regular, by Abhyankar, for every irreducible component of the exceptional set, the geometric generic fiber of the component over its image is a ruled variety. By construction, this fiber immerses into a fiber of $\mathcal{B}\to S$. Since these fibers are Abelian varieties, they admit no nonconstant morphism from a genus $0$ curve. Thus, $\mathcal{A}'\to \mathcal{A}$ is an isomorphism. The composition of the projection $\text{pr}_{\mathcal{B}}:\mathcal{A}'\to \mathcal{B}$ and the inverse of this isomorphism is an $S$-morphism $\phi:\mathcal{A}\to \mathcal{B}$ extending $K$. Thus, $\text{Hom}_S(\mathcal{A},\mathcal{B})$ has proper components over $S$.

The Rigidity Lemma from "Geometric Invariant Theory" implies that the morphism $\text{Hom}_S(\mathcal{A},\mathcal{B})\to S$ is unramified. To prove this, it suffices to prove that for every Artin local scheme $\text{Spec}(C)$ over $S$, for every pair of $C$-morphisms of the pullbacks, $$\phi,\psi:\mathcal{A}_C\to \mathcal{B}_C,$$ both of which map the identity section to the identity section, if the restrictions of $\phi$ and $\psi$ over $\text{Spec}(C/\mathfrak{m}_C)$ are equal, then $\phi$ equals $\psi$. In other words, for the difference morphism $\delta = \phi - \psi$, if $\delta_{C/\mathfrak{m}_C}$ equals $0$, then $\delta$ equals $0$. By the Ridigity Lemma, Proposition 6.1(1), p. 115 of "Geometric Invariant Theory", $\delta$ is a composition of the $C$-morphism $\mathcal{A}_C \to \text{Spec}(C)$ and a section $\sigma:\text{Spec}(C)\to \mathcal{B}_C$ of $\mathcal{B}$ over $\text{Spec}(C)$. Since $\delta$ sends the identity section to the identity section, $\sigma$ is the identity section. Thus, $\delta$ is the constant morphism to the identity section, i.e., $\phi$ equals $\psi$. Therefore $\text{Hom}_S(\mathcal{A},\mathcal{B})$ is unramified over $S$.

Since the components are both proper and unramified, then they are finite. For a non-unibranch varieties $S$, there can exist a finite, unramified morphism $S'\to S$ that is not étale. For instance, if $S$ is a nodal plane curve, then the normalization will be finite and unramified, but not étale. However, if $S$ is normal, then every finite, unramified, dominant morphism to $S$ is étale. Therefore, every irreducible component of $\text{Hom}_S(\mathcal{A},\mathcal{B})$ that dominates $S$ is finite and étale over $S$.