If $B$ is a complete boolean algebra without atoms (seen as a locale), then the categories $Sh(B,Ab)$ or $Sh(B,R\text{-Mod})$ are Grothendieck abelian categories that have no simple objects, in fact no indecomposable objects.
Here is the proof:
Assume $F$ is any object in $Sh(B,Ab)$ then for any element $U \in B$ we can construct the restriction $F|_U$ defined by $F|_U = i_* i^* U$ where $i:U \to X$ is the inclusion. or more explicitly $F|_U (V) = F(U \cap V)$, and we can always write that $F = F|_U \oplus F|_{\neg U}$.
Do if $F$ is simple (or just indecomposable) it follows that for every $U \in B$ we have either $F|_U = 0 $ or $F|_{\neg U} =0$.
From this we can deduce that if $F$ is an indecomposable object, then if we define $P = \{V \in B| F|_V \neq 0 \}$ we can check that $P$ is a point of $B$
Indeed, $1 \in P$ because $F$ is non-trivial, if $\bigcup U_i \in P$ then $F$ has to be non trivial on at least one of the $U_i$. and if $F$ is non-trivial on $V$ and on $W$, then $F|_{\neg V} = 0$, so $F$ has to be concentrated on $W \cap V$. hence $W \cap V \in P$.
This concludes the proof as points of complete boolean algebra corresponds to atoms, and we assume there was no atoms in $B$.