First question: no: consider e.g. $G:=\left[\begin{matrix} 0 & 1 \\0 & 0 \end{matrix}\right]$ and $G_0:=\left[\begin{matrix} 1 & 1 \\0 & 0 \end{matrix}\right]$ on $\mathbb{R}^2$.
Second question: a normal operator $G$ is unitarily equivalent to a multiplication operator by a measurable function $f$ on some $L^2$ space; $G_0$ is then the multiplication operator corresponding to the function $f+\chi_{\{f=0\}}$. So $G_0$ is invertible if and only if $\sigma(G)$ is bounded and $0$ is isolated in it.