At the best of my possibilities, my question should sound like this. Assume we have a category $C$ in which objects are sets (so I may speak about surjectivity and injectivity of morphisms in the elementary sense) and with small limits and assume we have a pair of morphisms $f,g:a\to b$ in $C$ admitting a common retraction $r:b\to a$. Assume also that they admit a coequalizer $p:b \to c$ and consider the kernel pair $b\times_cb$ of $p$. Then it is clear that there exists a unique morphism $f\times_cg:a\to b\times_cb$ given by the universal property of the kernel pair.
Question: Is $f\times_cg$ surjective?
I know that without the hypothesis on the common retraction, I could not expect this to be true. In Set, for example, the kernel pair of the coequalizer should be the smallest equivalence relation $R$ on $b$ such that $(f(\cdot),g(\cdot))\in R$. But what about in this more particular case?
Let consider your situation in a category with pullbacks. Then we have a commutative diagram:
Since $f$ and $g$ are sections, the comparison morphism ($f\times_cg$ in your notation) $k:a\to b\times_cb$ is a section as well. Consequently, $k$ is an epimorphism if and only if it's an isomorphism if and only if $f,g$ is the kernel pair of $p$.
Moreover, since kernel pairs are retractions, $f$ and $g$ are both sections and retractions, hence isomorphisms. Then $p$ is an isomorphism as well and $f=g$.