In $\mathbf{Rng}$, the inclusion $\mathbb{Z} \hookrightarrow \mathbb{Q}$ is monic and epic but not an isomorphism. If we slightly strengthen the requirements, however, we obtain:

Theorem: A map is an isomorphism if and only if it is monic and \underline{regular} epic.

Proof: Assume $m : B \rightarrow C$ is monic and regular epic. Then it is the coequalizer of some parallel pair $A \xrightarrow{f,g} B$. This means that $m \circ f = m \circ g$, but since $m$ is monic, $f = g$. Since this is the case, $\id_B$ equalizes $f$ and $g$, so we obtain a unique map $\varphi : C \rightarrow B$ such that $\varphi \circ m = \id_B$. Then, $m \circ \varphi \circ m = m = \id_C \circ m$. Since $m$ is epic (regular epic $\Rightarrow$ epic), we have $m \circ \varphi = \id_C$. So, $m$ is an isomorphism.

Suppose $m$ is an isomorphism. It's trivially monic, since given $m \circ f = m \circ g$, we can left-compose by $m^{-1}$ to obtain $f = g$, and similarly for epic. To show it is regular epic, we must exhibit it as the coequalizer of a parallel pair. Take the parallel pair $C \xrightarrow{m^{-1}, m^{-1}} B$. Clearly $m$ coequalizes the pair. Now suppose $B \xrightarrow{f} D$ coequalizes the pair (any map does, trivially). Define $\varphi : C \rightarrow D$ as $\varphi = f \circ m^{-1}$. Clearly this makes the appropriate triangle commute, since $f \circ m^{-1} \circ m = f$. Suppose there were another map $\phi$ with $\phi \circ m = f$. Right compose with $m^{-1}$ to get $\phi = f \circ m^{-1} = \varphi$. So $\varphi$ is the unique such map and $(m, C)$ is a coequalizer.