Yes, and there are more general results available. E.g. in the paper of H. H. Storrer, Epimorphic Extensions of Non-Commutative Rings, at the bottom of p. 74 there are references given for the following fact: Among those rings $R$ such that any epic monomorphism of (unital, associative) rings $\phi: R \rightarrow S$ is an isomorphism, there are the von Neumann regular and the self-injective rings. Compare also the first paragraphs of chapter XI in Bo Stenström's book Rings of Quotients.

And another proof in the vein of Martin's and Ralph's: If $0 \neq \alpha: R \rightarrow S$ is an epimorphism, every $R$-linear endomorphism of $S$ (as left $R$-module, say) would have to be $S$-linear. (In fact, $\alpha_*: S-Mod \rightarrow R-Mod$ being full is another equivalent criterion for $\alpha$ being an epi). Now as $R$ is a skew field, $\alpha(R)$ is a direct summand in $S$. If it has a non-trivial complement, we can certainly define non-identical $R$-linear endomorphisms of $S$ whose restriction to $\alpha(R)$ are identical; but any $S$-linear endomorphism of $S$ is determined by what it does on $1_S = \alpha(1_R)$. So $S = \alpha(R)$.

Yes, and there are more general results available. E.g. in the paper of H. H. Storrer, Epimorphic Extensions of Non-Commutative Rings, at the bottom of p. 74 there are references given for the following fact: Among those rings $R$ such that any epic monomorphism of (unital, associative) rings $\phi: R \rightarrow S$ is an isomorphism, there are the von Neumann regular rings and the self-injective rings. Compare also the first paragraphs of chapter XI in Bo Stenström's book Rings of Quotients.

And another proof in the vein of Martin's and Ralph's: If $0 \neq \alpha: R \rightarrow S$ is an epimorphism, every $R$-linear endomorphism of $S$ (as left $R$-module, say) has to be $S$-linear. (In fact, $\alpha_*: S-Mod \rightarrow R-Mod$ being full is another equivalent criterion for $\alpha$ being an epi). Now as $R$ is a skew field, $\alpha(R)$ is a direct summand in $S$. If it had a non-trivial complement, we could certainly define non-identical $R$-linear endomorphisms of $S$ whose restrictions to $\alpha(R)$ are identical; but any $S$-linear endomorphism of $S$ is determined by what it does on $1_S = \alpha(1_R)$. So $S = \alpha(R)$.