Perhaps a concrete and simple example is helpful to develop intuition. Take the Laplacian $-d^2/dx^2$ on $\mathbb{R}$. The spectrum is continuous and unbounded, $E(k)=k^2$, $k\in\mathbb{R}$.
Now discretize the $x$ variable on $\mathbb{Z}$. The spectrum remains continuous, but becomes bounded, $\tilde{E}(k)=2-2\cos k$, $k\in(-\pi,\pi]$. Notice that for small $k$ the two spectra coincide, $\tilde{E}(k)=E(k)+{\rm order}(k^4)$. Small $k$ means large wave lengths, and the discretization is not noticed if the wave length is larger than the spacing of the discretization.
You ask: "when we discretize the eigenvalues numerically, what do we expect to see?" The discrete spectrum appears not because of the discretization, but because numerically you will need to consider a finite system, rather than an infinite system. So you will restrict $x$ to some finite interval $(-N,N]$ of $\mathbb{Z}$ and you will need to introduce boundary conditions at the end points. Periodic boundary conditions are convenient, then the spectrum becomes $E_n=2-2\cos(n\pi/N)$, $-N<n\leq N$. For other boundary conditions the formula is different, but the spectrum remains discrete.
This is for a self-adjoint operator with a real spectrum, but I think the same applies to a non-self-adjoint operator: numerically you find a discrete spectrum not because of the discretization, but because of the finite system.
