According to Corollary 1.2(3) of the paper Silver: Noncommutative Localizations and Applications. J. of Alg. 7(1964), 44-67:
If $R$ is a (commutative) field and $\alpha: R \to S$ an epimorphism in the category of rings, then $\alpha$ is an isomorphism.
Question: Is $\alpha$ also an isomorphism if $R$ is not a field but just a division ring ?
For convenience, I repeat the short proof from Silver (which doesn't seem to work for division rings): First it is noted (Prop. 1.1) that a homomorphism $\alpha: R \to S$ of (not necessarily commutative) rings with identity is epi iff multiplication $S \otimes_R S \to S$ is an isomorphism (of abelian groups). Then:
"For $x\in S$, consider the subring $R[x]$ of $S$ generated by $R$ and $x$. Since $R$ is a field, one can easily see that $R \to R[x]$ is an epimorphism using 1.1. If $x$ is transcendental over $R$, then $\beta: R[x] \to R[x]$ defined by $\beta(f)=f(0)$ agrees on $R$ with the identity map of $R[x]$. So $x$ cannot be transcendental over $R$ by definition of an epimorphism. Finally, if $[R[x]:R] < \infty$, then by 1.1, $[R[x]:R]^2=[R[x]:R]$, so $[R[x]:R]=1$ and $x \in R$. Hence $\alpha$ is an isomorphism as desired."