The action of $S$ on $X$ induces an action of $BS$ on $BX$. One way to see it is to do it at the level of $p$ simplices and construct a map $B_pS\times B_pX\to B_pX$ for all $p$.

This action induces an action map $H_0(BS)\otimes H_p(BX)\to H_p(BX)$ making $H_p(BX)$ into an $H_0(BS)$-module. But $H_0(BS)=\mathbb{Z}[\pi_0(BS)]$, therefore you have an action of the group algebra $\mathbb{Z}[\pi_0(BS)]$ on $H_p(BX)$ or equivalently an action of the group $\pi_0(S)$ on $H_p(BX)$.

The action of $S$ on $X$ induces an action of $BS$ on $BX$. One way to see it is to do it at the level of $p$ simplices and construct a map $B_pS\times B_pX\to B_pX$ for all $p$. For any category $C$, the $p$ simplices $B_pC$ are the functors $[p]\to C$. Since $S$ acts on $X$, we have a functor $S\times X\to X$. Mapping $[p]$ to this map, we construct a map: $B_pS\times B_pX\to B_p X$ (note that $Fun([p],-)$ preserves product). It is formal to check that this map induce a map of simplicial sets $B_\bullet(S)\times B_\bullet(X)\to B_\bullet(X)$. If you would rather work with topological space, you can use the fact that geometric realization is product preserving and you can turn this map into a map of topological spaces.

This action induces an action map $H_0(BS)\otimes H_p(BX)\to H_p(BX)$ making $H_p(BX)$ into an $H_0(BS)$-module. But $H_0(BS)=\mathbb{Z}[\pi_0(BS)]$, therefore you have an action of the group algebra $\mathbb{Z}[\pi_0(BS)]$ on $H_p(BX)$ or equivalently an action of the group $\pi_0(S)$ on $H_p(BX)$.