This is true in general Banach spaces $X$ (see Dunford-Scwartz vol I for the general theory I use)  . The assumption $A_0S=0$ says that $0$ is a simple pole of the resolvent so that $X=Ker (A_0) \oplus Im (A_0)$ and the splitting is given by the projection $S$, that is $Ker (A_0)=S(X)$, $Im(A_0)=Ker (S)$. Then $A_0$ is invertible from $Im(A_0)$ into itself and $S$ is the identity on $Ker (A_0)$ and identically zero on $Im (A_0)$ and then $A_0+S$ is invertible.

This is true in general Banach spaces $X$ (see Dunford-Schwartz vol I for the general theory I use). The assumption $A_0S=0$ says that $0$ is a simple pole of the resolvent so that $X=Ker (A_0) \oplus Im (A_0)$ and the splitting is given by the projection $S$, that is $Ker (A_0)=S(X)$, $Im(A_0)=Ker (S)$. Then $A_0$ is invertible from $Im(A_0)$ into itself and $S$ is the identity on $Ker (A_0)$ and identically zero on $Im (A_0)$ and then $A_0+S$ is invertible.