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 (meaning "linear homeomorphism") if and only if $\sigma(G)$ is bounded and $0$ is isolated in it.
edit: As pointed out by Christian Rempling, the requirement was just $0\in\rho(G_0)$, that is, $G_0$ injective and $G_0^{-1}\in B(H)$, which is translated in $0$ being isolated in $\sigma(G)$. This is true if for some $\lambda_0$.$(\lambda_0-G)^{-1}$ is a compact operator (for then $\sigma(G)$ is a discrete unbounded set).