Consider a densely defined linear operator $A$ on a Hilbert space $H$ over $\mathbb{C}$. That is, $A$ is not necessarily a bounded linear operator. Suppose that $A$ has a compact inverse mapping $B$. It is well-known that 1.) the spectrum of a compact linear operator is discrete, 2.) every non-zero element of $\sigma(B)$ is an eigenvalue of $B$, 3.) every eigenvector $v$ of $B$ with eigenvalue $\lambda$ is an eigenvector of $A$ with eigenvalue $\lambda^{-1}$.
Do we then know that the spectrum of $A$ is also discrete? In the case that $A$ was a compact self-adjoint operator, one way to deduce that if $\lambda\neq 0$ is not non-eigenvalue of $A$ is to first take eigenvectors $\left(u_k\right)_{k\in\mathbb{N}}$ of $A$ and to show that $(A - \lambda I)$ has a bounded inverse by $(A - \lambda I)^{-1}u = \lambda^{-1}Pu + \sum_{k\in\mathbb{N}}(\lambda - \lambda_k)^{-1}\left<u,u_k\right>u_k$ with $Pu = u - \sum_{k\in\mathbb{N}}\left<u,u_k\right>u_k$. And to my naïve eye there does not seem to be anything preventing us to using similar construction in the case that $A$ is unbounded, but is at least symmetric. Although to be honest I don't see why the symmetric property is important in this case.
So to reiterate, if $A$ is just a densely defined linear operator over a Hilbert space $H$ with a compact inverse, is $A$'s spectrum also discrete and consists solely of the eigenvalues of $A$? And if so, how can one show this?
The way of construction of the inverse presented in OP fails if $A$ is not symmetric. For example let $\mathcal{H}=L^2(0,1)$ and $Af=f',$ with $D(A)=\{f\in C^1[0,1]\,:\, f(0)=0\}.$ The inverse is given by $$(Bg)(x)=\int\limits_0^xg(t)\,dt$$ The operator $B$ is compact but has no eigenvalues.
For a symmetric operator $A$ we may encounter a problem as well. Let $\{e_n\}_{n=1}^\infty$ denote an orthonormal basis of $\mathcal{H}$ and $$D(A)={\rm span}\, \{e_n-e_{n+1}\}_{n=1}^\infty,\quad Ax=\sum_{k=1}^\infty k\langle x,e_k\rangle e_k$$ Then $D(A)$ is dense and $$By=\sum_{k=1}^\infty {1\over k}\langle y,e_k\rangle e_k$$ Hence $B$ is a compact operator with eigenvectors $e_k,$ which are not eigenvectors of $A,$ as $e_k\notin D(A).$
We may extend the operator $A$ to $\tilde{A}$ by setting $D(\tilde{A})={\rm span}\, \{e_n\}_{n=1}^\infty,$ and the problem dissappears.