Let $A_0$ be a bounded linear operator on a Hilbert space $H$. Suppose $0$ is an isolated point of the spectrum of $A_0$. Let $S$ be the corresponding Riesz projection, namely, $$S = -\frac{1}{2\pi i} \int_{C_\varepsilon} (A_0 - \lambda)^{-1}d\lambda,$$ where $C_\varepsilon$ is a circle of small radius $\varepsilon$, centered at $0$, contained entirely in the resolvent set of $A_0$. By Cauchy's Theorem, the definition of $S$ is independent of such $\varepsilon$. Assume also that $$A_0 S = 0.$$
Using the work of this previous post, we have $\ker S = \text{ran }A_0$ and $\ker A_0 = \text{ran }S$.
I would like to know whether $A_0 + S$ has a bounded inverse. Note in particular that I am not assuming $A_0$ is self-adjoint.
By the open mapping theorem, it suffices to show that $A_0 + S$ is bijective on $H$.
For injectivity, suppose $(A_0 + S) \psi = 0$. We want to see $\psi =0$, and we make use of $\ker S = \text{ran } A_0$ and $\ker A_0 = \text{ran } S $, which we showed above: $S\psi = -A_0 \psi \in \ker S$. Thus $S\psi = S^2 \psi = 0$, hence $A_0 \psi = 0$, so $\psi \in \ker A_0 = \text{ran } S$. Therefore $\psi = S\psi = 0$.
I am stuck showing surjectivity (if indeed it is true). The best I can do is as follows, which is inspired by the older post. If $\psi \in H$, consider the element $\phi = \tfrac{1}{2\pi i} \int_{C_\varepsilon(\lambda_0)} \lambda^{-1} (A_0 - \lambda)^{-1} \psi d\lambda$. We compute \begin{equation*} \begin{split} A_0\phi &= \frac{1}{2\pi i} \int_{C_\varepsilon(\lambda_0)} \lambda^{-1} \psi d\lambda \\ &+ \frac{1}{2\pi i} \int_{C_\varepsilon(\lambda_0)}(A_0 - \lambda)^{-1} \psi d\lambda \\ &=\psi - S\psi. \end{split} \end{equation*} If it happens that $S\psi = S\phi$, we are done, but I couldn't verify this.