If I have a maximum principle of the form: If $\psi\geq 0$ of the boundary. Then $L\psi\geq 0$ on the domain $D$ implies $\psi\geq 0$ on the closure of $D$, for some linear operator $L$ defined in terms of $\Delta$.
I want to use this to show that if $|Lz|\leq C$ then $|z|\leq C$.
I assume, the operator $L$ is a second-order elliptic differential operator. Then the conditions you gave are not sufficient to prove the desired result.
(1) Let $e\equiv 1$ on $D$. If you assume $Le\ge \sigma$, $\sigma \ge1$, then you can prove it:
Assume $|Lz|\le C$ on $D$, $|z|\le C$ on $\partial D$. Then $z-C\le 0$ and $$ L(z-C) = Lz- CLe\le C - C\sigma =C(1-\sigma)\le 0, $$ which implies $z\le C$ on $D$.
(2) If you are just interested in proving the boundedness of $z$, then you might look into the literature about weak maximum principle (eg Gilbarg/Trudinger, Section 8.5). Under suitable assumptions, one gets the estimate $$ \|u\|_{L^\infty(D)}\le \|u\|_{L^\infty(\partial D)} + c\|Lu\|_{W^{-1,q}(\Omega)} $$ with $q>n$, $n$ spatial dimension; $c$ depends on $\Omega$, $n$, and the coefficient functions of $L$.