It is easy to see that in the category $\sf Rng$ of rings, the equalizer of two homomorphisms $f,g\colon R \to S$ is just the inclusion $\ker(f_1-f_2) \hookrightarrow R$. Consider then the category $\sf Ring$ of rings with $1$ and ring homomorphisms mapping $1$ to $1$. Do we still have equalizers in this category? I feel the answer should be no, especially if the morphisms are non-trivial, since then the natural candidate $\ker(f_1-f_2)$ will not be an object of ${\rm Ring}$ (to make it worse, $1 \in \ker(f_1-f_2)$ forces actually $f_1=f_2$).
Is there any way around this?
Any algebraic category is complete, and the carrier of the limit is the limit of the carriers. Since adding a unit corresponds to adding a constant to the theory of rngs and some equational laws, the category of rings with unit is an algebraic category. The equalizer of two rings is then the equalizer of their carrier sets with the operations restricted to the subset. (You should verify that they are closed.)
Kernels are usually used when a category has a zero object (and thus zero morphisms). This is not the case for $\mathbf{Ring}$. $\mathbf{Ring}$ similarly doesn't admit an abelian group structure on its homsets (i.e. it isn't $\mathbf{Ab}$-enriched). At least not in the obvious way. The difference of two (unit preserving) ring homomorphisms isn't a unit preserving ring homomorphism because $(f-g)(1)=f(1)-g(1)=0\neq 1$. Trying to reduce equalizers to kernels doesn't make sense in this context.