Some assumptions are needed. I suppose $\phi \in L^{\infty}(\mu)$ and that $\mu$ is just a semi-finite measure.
Item 1: Let $N = \{ x:\phi\left(x\right)=0\}$.
$(\Rightarrow)$ If $\mu(N)>0$, then (since $\mu$ is a semi-finite) there is $S \subseteq N$ such that $0<\mu(S)<+\infty$.
Then $\mathbf{1}_S \in L^{2}(\mu)$, $\mathbf{1}_S \ne 0$ and
$ M_{\phi} \mathbf{1}_S = \phi \mathbf{1}_S =0 $.
So $ \mathbf{1}_S \in \ker M_{\phi}$ and $\ker M_{\phi} \ne 0$.
So we have proved that, if $\ker M_{\phi} = 0$, then $\mu(N)=0$.
$(\Leftarrow)$ If $\ker M_{\phi} \ne 0$, then there is $f \in L^{2}(\mu)$ such that $f \ne 0$ and $ \phi f = M_{\phi} f =0$. Let $C= \{ x: f(x) \ne0\}$.
Since $f \ne 0$, we have that $\mu(C)>0$. Since $\phi f = 0$, we have that $\phi(x) = 0$, for almost all $x \in C$. So, there is $D \subseteq C$ such that $\mu(D) =\mu(C)>0$ and $\phi(x) = 0$, for all $x \in D$. But then $D \subseteq N$ and $\mu(N) >0$.
So we have proved that, if $\mu(N)=0$, then $\ker M_{\phi} = 0$.
Item 2: Let $X$ be the space, $N = \{ x:\phi\left(x\right)=0\}$ and $N^c = X\setminus N$.
Let us prove that
$\mbox{ran}M_{\phi}$ be closed if and only if there is $c>0$ such that $|\phi(x)|>c$, for almost all $x \in N^c$.
$(\Leftarrow)$ Suppose there is $c>0$ such that $|\phi(x)|>c$, for almost all $x \in N^c$.
Given any $g \in \overline {\mbox{ran}M_{\phi} } $, let $f_n$ be a sequence such that $\phi f_n$ converges to $g$. So $\phi f_n$ is a Cauchy sequence. But we have that
$$ \|f_m|_{N^c}-f_n|_{N^c}\| \leq c^{-1} \|\phi f_m|_{N^c} - \phi f_n|_{N^c} \|$$
So, $f_n|_{N^c}$ is also a Cauchy sequence. So, there is $f \in L^{2}(\mu)$ such that $f_n|_{N^c}$ converges to $f|_{N^c}$ (we may define $f|_N=0$). Since $M_{\phi}$ is continuous, we have that
$$M_{\phi} f= M_{\phi} f|_{N^c} + M_{\phi} f|_{N} =M_{\phi} f|_{N^c}= \lim_n M_{\phi} (f_n|_{N^c})= \lim_n M_{\phi} f_n = \lim_n \phi f_n =g$$
So $g \in \mbox{ran}M_{\phi} $. So $\mbox{ran}M_{\phi}$ is closed.
$(\Rightarrow)$
We clearly have that $ L^{2} (N^c, \mu)$ is a close subspace of $ L^{2} (X, \mu)$.
Moreover, it easy to see that $M_{\phi}$ is bijective (and continuous) from $ L^{2} (N^c, \mu)$ onto $\mbox{ran}M_{\phi}$.
So, if $\mbox{ran}M_{\phi}$ is closed, then $\mbox{ran}M_{\phi}$ is itself a Banach space, and then by the Inverse Mapping Theorem, we have that $M_{\phi}^{-1}$ is continuous and so it is bounded. So,
there is $c>0$, such that
$$\| M_{\phi} f\| \geq c \|f\|$$
for all $f \in L^{2} (N^c, \mu)$. It follows that $|\phi(x)|>c$, for almost all $x \in N^c$.
Remark: we used the following facts
For any function $f \in L^2(X, \mu)$, $ f = f\mathbf{1}_{N} + f\mathbf{1}_{N^c}$ . So $M_\phi f = M_\phi (f\mathbf{1}_{N}) + M_\phi (f\mathbf{1}_{N^c})=0+ M_\phi (f\mathbf{1}_{N^c})= M_\phi(f|_{N^c}) $.
Suppose $g \in L^2(N^c, \mu)$. We can consider the function $g$ to be a functions defined on $X$ by considering $g(x)=0$ for $x \in N$. It is the canonical inclusion of $L^2(N^c, \mu)$ into $L^2(X, \mu)$. It is easy to see that, under this inclusion, $L^2(N^c, \mu)$ is a closed subspace of $L^2(X, \mu)$. In fact, let $g_n$ be a sequence of functions in $L^2(N^c, \mu)$ converging to $g\in L^2(X, \mu)$. Then, there is a subsequence $g_{n_k}$ that converges almost everywhere to $g$. So $g(x)=0$, for almost every $x \in N$. So $g = g \mathbf{1}_{N^c}$, which means that $g \in L^2(N^c, \mu)$.
$M_\phi$ restricted to the subspace $L^2(N^c, \mu)$ is injective. Just note that $\{x \in N^c : \phi(x)=0\}= \emptyset$ and apply item 1 to conclude that $\ker(M_\phi|_{L^2(N^c, \mu)})=0$.
