Let $\mathbb C$ be a small regular category. Let $J$ be the Grothendieck topology generated by coverings $\{U'\twoheadrightarrow U\}$ consisting of precisely the regular epimorphisms in $\mathbb C$. The topology is subcanonical, and the Yoneda embedding $\mathbb C \to \mathsf{Sh}(\mathbb C,J)$ preserves regular epimorphisms, images and of course finite limits. Does it preserve coequalizers?
Edit. Here is what I tried. I have a bunch of diagrams. I start with a coequalizer in $\mathbb C$ this is the diagram on the left upper corner. The diagram below it explains what $k_0$, $k_1$ and $j$ are. The pair $(k_0,k_1)$ is the kernel pair of $q$. I send my coequalizer diagram into the sheaf category via Yoneda and take a test sheaf $F$ with a test map $b$ into $F$ which equalizes $\mathbf yf$ and $\mathbf y g$. I like to factor $b$ uniquely through $\mathbf yq$. By the Yoneda-lemma I see that my factorisation problem is equivalent to the one sketched in the diagram below! I need to show that any element $b$ of $FB$, which gets send to the same element by the two maps into $FA$, already lies in $FC$. I know that by assumption ($F$ is a sheaf in the regular topology) the upper fork is an equalizer diagram in the category of Sets. So if I can show that the element $b$ is also equalised by the two maps into $F(B\times_CB)$, then I am done. But unfortunately I can not just conclude that this is the case, because $Fj: F(B\times_CB) \to FA$ points in the wrong direction :( So I am stuck.
Edit edit. If the embedding does not preserve coequalisers, then I would be also satisfied by a weaker result. Assume $\mathbb C$ is additionally effective regular and additive (i.e. abelian). Then the embedding $\mathbb C \to \mathsf{Sh}(\mathbb C,J)$ embeds the abelian category into the subcategory of abelian group objects in the topos $\mathsf{Sh}(\mathbb C,J)$. This embedding preserves the zero object and the additive structure and kernels, but does it also preserve cokernels?