Let $G$ be a normal operator with compact resolvent on a Hilbert space $H$ such that ${\rm ker}(G) \neq {0}$. Further let $P$ be the orthogonal projection onto ${\rm ker}(G)$, and let $G_{0}:=G+P$.
What is the relation between the spectra of $G$ and $G_{0}$?
If $G$ is normal (and you don't care it has compact resolvent or not), then $G_0 = G +P$ is $f(G)$ where $f$ is the measurable function that send $0$ to $1$ and is the identity on other value.
If I remember correctly when $G$ is normal the spetrum of $f(G)$ is the closure of the $f(spec G)$, so it gives the answer in the comment: you add one to the spectrum (because you assumed that $0$ was in the spectrum) and you remove $0$ if it is isolated (if $0$ is not isolated then you just add $1$).
Anyway, this can be proved and made more explicit using the spectral theorem: the spectral theorem say that your Hilbert space is isomorphic to $L^2(K)$ for some sigma finite measured space $K$ with $G$ identified with the multiplication by some measurable function $g:K \rightarrow \mathbb{C}$ on $L^2(K)$ (for some measure on $K$). For such an operator, The spectrum of this operator is essentially the closure of the image of $g$ or more precisely, the set of complex numbers $\alpha$ such that for every $\epsilon$, $g$ is $\epsilon$-close to $\alpha$ on set of non-zero measure.
The operation you are describing just replace $g$ by the function $g_0$ which is $1$ if $g=0$ and $g$ otherwise, indeed $P$ is the multiplication by the function that is $1$ if $g=0$ and $0$ otherwise. As the kernel is non-empty it adds one to the spectrum, the spectrum outside $0$ and $1$ is left unchanged and the case of $0$ only depends on if $0$ is isolated or not:
If $0$ is isolated, then $g_0$ is ``never'' (more precesely, there is an $\epsilon >0$ such that $g_0^{-1} B(0,\epsilon)$ has zero measure) close to $0$ hence $0$ disapears from the spectrum, if $0$ was not isolated then $g$ take arbitrary small values different from $0$, hence $g_0$ too and $0$ stay in the spectrum.
