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2 Noted a small mistake.

Let's make several reductions. First, the condition implies $f$ is injective, so letting $K=\operatorname{Frac}(R)$, we have that $f: K\to S'$ is an epimorphism, letting $S'$ be the localization of $S$ at the zero ideal of $R$; it is easy to check that this is an epimorphism via the universal property of localization.

But as George Lowther points out, an epimorphism from a field to an integral domain must be surjective. To see this, assume to the contrary that $K\to S'$ is not surjective; thus $K'=\operatorname{Frac}(S')$ is not equal to $K$. But $S'\to K'$ is an epimorphism, so composing, we've reduced to the case of a map between two fields $f: K\to K'$. But $K'$ admits many embeddings into (say) its algebraic closure which agree on $K$, contradicting that $f$ was an epimorphism.

EDIT: As the commenters point out, the algebraic closure doesn't quite work in the case $K'/K$ is purely inseparable, but this case is not difficult; see George Lowther's answer for an easy general argument.

1

Let's make several reductions. First, the condition implies $f$ is injective, so letting $K=\operatorname{Frac}(R)$, we have that $f: K\to S'$ is an epimorphism, letting $S'$ be the localization of $S$ at the zero ideal of $R$; it is easy to check that this is an epimorphism via the universal property of localization.

But as George Lowther points out, an epimorphism from a field to an integral domain must be surjective. To see this, assume to the contrary that $K\to S'$ is not surjective; thus $K'=\operatorname{Frac}(S')$ is not equal to $K$. But $S'\to K'$ is an epimorphism, so composing, we've reduced to the case of a map between two fields $f: K\to K'$. But $K'$ admits many embeddings into (say) its algebraic closure which agree on $K$, contradicting that $f$ was an epimorphism.