I think I have an answer when $(\Omega,\mathcal F,\mu)$ is a $\sigma$-finite measure space.
For $1\leq p<\infty$, $M_m$ is dispersive if and only if we have $\Re m(x)\leq 0$ for almost every $x$.
We show the result when $(\Omega,\mathcal F,\mu)$ is finite. We assume that $M_m$ is dispersive. We put for $\lambda\in\mathbb Q_+^*$ and $n\in\mathbb N^*$, $A_n^{\lambda}:=\left\{x\in \Omega\mid|\lambda -m(x)|^p-|\lambda|^p\leq -\frac 1n\right\}$. We have for each $f\in L^p(\Omega)$ and $\lambda\in\mathbb Q_+^*$ $\int_{\Omega}|\lambda-m(x)|^p|f(x)|^pd\mu x\geq \lambda^p\int_{\Omega}|f(x)|^pd\mu,$ and in particular, for $f=\mathbf 1_{A_n^{\lambda}}$, which is integrable $0=\int_{A_n^{\lambda}}(|\lambda-m(x)|^p-|\lambda|^p)d\mu\leq -\frac 1n\mu\left(A_n^{\lambda}\right),$ hence $\mu\left(A_n^{\lambda}\right)=0$ and the set $N:=\bigcup_{n>0,\lambda\in\mathbb Q^*_+}A_n^{\lambda}$ has measure $0$. Hence we have for each $x\in \omega\setminus N$ and $\lambda\in\mathbb Q_+^*$, $|\lambda-m(x)|\geq \lambda$ and by continuity this inequality is true for $\lambda \in\mathbb R$. Writing $m(x)=f(x)+ig(x)$ where $f$ and $g$ are real functions, we get $(\lambda-f(x))^2+g(x)^2\geq \lambda^2$ hence $-2\lambda f(x)+f(x)^2+g(x)^2\geq 0$ and $f(x)\leq \frac{f(x)^2+g(x)^2}{2\lambda}$ so $f(x)\leq 0$ for each $x\in \Omega\setminus N$.
Conversely, we assume that $f(x)\leq 0$ for almost every $x$. We have for each $\lambda>0$ and for almost every $x$|\lambda-f(x)-ig(x)|^2-\lambda^2=-2\lambda f(x)+f(x)^2+g(x)^2\geq 0 which yields the inequality $\lVert (\lambda I-M_m)f\rVert_p\geq \lVert f\rVert_p$ for all $f\in L^p(\Omega).
To jump from the finite case to the \sigma$-finite one, write $\Omega$ as a countable disjoint union of set of finite measure $\Omega_n$, and apply what was done to this case to get that we should have almost everywhere on $\Omega_n$, $\Re m(x)\leq 0$, hence almost everywhere on $\Omega$.