Suppose $R\subseteq S$ is a flat extension of rings and $A/R$, $B/S$ are flat Hopf algebras. Let $\varphi:A\otimes_R S\to B$ be a surjective $S$-Hopf algebra homomorphism. When is it the case that $\varphi(A)=\varphi(A\otimes 1)$ is a Hopf algebra over $R$ such that $\varphi(A)\otimes_R S=B$ as $S$-Hopf algebras?
(Rephrasing: suppose $Y\to X$ is a flat morphism of schemes, $G/Y$ and $H/X$ are flat affine group schemes, and we have a closed inclusion $G\to H\times_X Y$ of group schemes over $Y$. When does this come from pulling back some inclusion $\mathcal{G}\to H$? My main motivation for considering this question is the case $R$ is a Dedekind domain, $S=\mathrm{Frac}(R)$, and $H=\mathrm{GL}_n$, to obtain integral models from faithful representations.)
If we suppose $\varphi(A)/R$ is flat and $S$ is a localization of $R$ (so that $-\otimes_R-=-\otimes_S-$), then I believe this follows by using the following fact three times to construct the structure maps: in any Abelian category, a diagram of the form
has a unique solution. Commutativity of the diagrams necessary to make this a Hopf algebra then comes from, say, using flatness assumptions to embed everything into the corresponding commutative diagrams associated to $B$.
However, without the assumption that $S/R$ is a localization, this fails: for example, it is not too hard to construct counterexamples for $S/R$ a quadratic étale cover. This is not too surprising: for example, if $L/K$ is a finite field extension and $G/K$ is an affine algebraic group, not every subgroup of $G\otimes_K L$ necessarily comes from a subgroup of $G$. In situations like this, can we say anything?
Note by "ring" I mean commutative ring and by "Hopf algebra" I mean commutative Hopf algebra (but not necessarily with commutative comultiplication).