This identity holds for arbitrary cochains, not only for cocycles. (As I wrote, $a\in C^n(G,M)$.) As a consequence, in the particular case when $a\in Z^n(G,M)$, we get $sa-a-db=c=0$ so $sa=a+db$, which implies $s[a]=[a]$, so we get an alternative proof of the fact that $G$ acts trivially on $H^n(G,M)$. However, this is not the reason why I need this result. I need it for arbitrary cochains, not for cocycles. (And only for $n=1$ or $2$ and $G$ commutative.) I know how to prove it, but if it is already somewhere, I would rather quote it than write it down. (It's a reference request.)

Sorry, I'm new to this and I didn't know that posting the same thing on math.stackexchange and mathoverflow is considered cross-posting. I thought they are separate things for different audiences (math.stackexchange for general public, mathoverflow forresearchers).