Consider the operator defined by $$M_m = \{ (\varphi,\psi) \in L^p(\mu) \times L^p(\mu) \ \vert \ \psi = m\varphi \}.$$
I've been able to show that this operator is closed and densely defined on the subspace on the set $X = \{ \varphi \in L^p \ \vert \ m\varphi \in L^p \}$.
I am new to semigroup theory. How does one show that $M_m$ is the generator of the semigroup $T(t)f : = e^{mt}f$?
I think you need that $Re(m)\leq \omega$ for some $\omega \in \mathbb R$ and almost everywhere. Otherwise $T$ is not well defined on $L^p(\mu)$.
Let $A$ be the generator of $T$. For $f\in dom(A)$ you have $$Af = \lim_{t\to 0}\frac{e^{tm}f-f}{t}.$$ Now use that this implies that you find a sequence $t_k\to 0$ along which this limit exists pointwise almost everywhere to conclude that $Af(x) = m(x)f(x)$ almost everywhere and hence $Af = M_m f$ in $L^p$.
This shows $A\subset M_m$.
Now here is a general useful fact about unbounded operators: If $A\subset B$ and there is $\lambda\in \mathbb C$ such that $\lambda-A$ is surjective and $\lambda-B$ is injective, then $A=B$. I would advise you to show that and then use it to show $A=M_m$.
Remember that for a generator of a semigroup there is always a full half plane contained in the resolvent.