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 $$A_0 S = 0.$$
It's not too difficult to see that $\ker A_0 \subseteq \text{ran}\, S$ (it essentially follows from the fact that $(A_0 - \lambda)^{-1} = -\lambda^{-1}$ on $\ker A_0$). So additionally assuming $A_0 S = 0$ gives $\ker A_0 = \text{ran}\, S$. Moreover, since $A_0$ commutes with its resolvent, we have $SA_0 = A_0 S = 0$, so $\text{ran}\,A_0 \subseteq \ker S$.
I would like to show additionally that $\ker S \subseteq \text{ran}\, A_0$, thus the range of $A_0$ is closed.
I have not made much progress on this problem, but here is what I tried so far. If $\psi \in \ker S$, then
$$0 = S\psi = -\frac{A_0}{2\pi i} \int_{C_\varepsilon} (A_0 - \lambda)^{-1}d\lambda \psi +\frac{1}{2\pi i} \int_{C_\varepsilon} (A_0 - I) (A_0 - \lambda)^{-1}d\lambda \psi \\ = -\frac{A_0}{2\pi i} \int_{C_\varepsilon} (A_0 - \lambda)^{-1}d\lambda \psi +\frac{1}{2\pi i} \int_{C_\varepsilon} (\lambda - I) (A_0 - \lambda)^{-1}d\lambda \psi $$ So if we can manage to show $(2\pi i)^{-1} \int_{C_\varepsilon} (\lambda - I) (A_0 - \lambda)^{-1}d\lambda \psi = \psi$, we are done. This clearly has something to do with the residue of the resolvent at zero, but I don't know how to compute that.