Here, $1 \le p < \infty$
I thought that maybe - for $a \neq 0 $ a.e. in $\Omega$ - it is true, that
$$\mathcal{C}^{\infty}_0 \subset \mathcal{R}(M_a)$$
And because $\mathcal{C}^{\infty}_0$ is dense in $L_p (\Omega)$, then this would mean that the range of $M_a$ is dense too in $L_p (\Omega)$
But I don't know if we can argue like that here.
The next step in that exercise would be to show that $\mathcal{R}(M_a) = L_p (\Omega)$, if there exists a constant $c>0$ such that $|a(x)| \ge c$ a.e. in $\Omega$
That's why I doubt my solution would be the right one here
Under the assumption that $1\leq p<\infty$, If $M_a(L_p)$ is to be contained in $L_p$, then $a$ must be in $L_\infty$: Suppose not. Then there are infinitely many sets in the sequence $E_n=\{n\leq a<n+1\}$ that have positive measure, say $E_{n_k}$ $k\leq n_k<n_{k+1}$. Choose $A_{n_k}\subset E_{n_k}$ be positive and finite measure and define $$g(x)=\sum_k\frac{1}{k\,(m(A_{n_k}))^{1/p}}\mathbb{1}_{A_{n_k}}$$ Then $g\in L_p$, however $$\int|ag|^p\,dm=\sum_k\frac{1}{k^p m(A_{n_k})}\int_{A_{n_k}}|a|^p\,dm\geq\sum_k\mathbb{1}=\infty$$ Thus $a\in L_\infty$ and moreover, $$\|M_af\|_p\leq\|a\|_\infty\|f\|_p$$
If $M_a(L_p)=L_p$, then there is $c>0$ such that $c<|a|\leq\|a\|_\infty$ (and vice versa).
I outline the prove here and leave the details to the OP
Show that $m(|a|=0)=0$. If that were not the case, choose $E\subset\{|a|=0\}$ with positive finite measure. Define $g=\mathbb{1}_E$. Clearly $g\in L_p$, but there is no $f\in L_p$ such that $M_af=g$.
W.l.g. assumee $|a|>0$. Then $M_a$ is one to one. As $M_a$ is onto, and application of the open map theorem implies that $M_a$ has a bounded inverse. Notice that the inverse of $M_a$ is $M_{1/a}$. Then $1/a\in L_\infty$ (why?).