Reading Evans and this note after asking this question, I have been thinking about the estimates for interior/global regularity in Evans, 6.3.1, theorem 1, and thoerem 4 in 6.3.2, of the form $$\|u\|_{H^2(V)}\le C(\|f\|_{L^2(U)}+\|u\|_{L^2(U)})$$ and $$\|u\|_{H^2(U)}\le C(\|f\|_{L^2(U)}+\|u\|_{L^2(U)})$$ Sadly, it seems in the first estimate general $V(\subset\subset U)$ cannot go to up the boundary without the trase zero condition, as the note above or Why does Boundary $H^2$ regularity fail for trace non-zero functions? says.
But the counterexample by @levap in the above question makes me wonder if it is possible to get (not the fact that $\|u\|_{H^2(U)}$ is bounded itself, but) the global regularity inequality of the form above without the trace zero condition for elements of a finite dimensional subspace $X_N\subset H^2(U)$. Here, elements of $X_N$ do not necessarily vanish on the boundary. The boundary can be really smooth if that is needed.
The following is a motivation if you are interested: If this type of estimate is available it would be convenient for analysing approximation methods such as the finite element method. For example, try to approximate the solution of $-\Delta u =f$ in $\Omega$ with $u=0$ on $\partial \Omega$ with the finite element method with hat function, say (further smoothness of functions can be assumed). Suppose $\Omega$ is not polygon. Then, in general approximation space $X_N={\rm span}\{\text{spanned by hat functions}\}$ is not a subspace of the solution space $H^{\text{some order}}_0(\Omega)$, since $v\in X_N$ does not vanish on the boundary, but takes zeros only on the nodes on the boundary. This makes analysis tedious, as the error $u-u_N$ does not vanish on the boundary. (Googling "variational crime" will show you details.) If we have the above estimate for $H^2_0\cup X_N$, that would be super.
I haven't a clue where to start by the way...
