If a module $M$ is *simple*, that means its only submodules are $0$ and $M$.
If $M = 0$ there is nothing to show, so pick a non-zero $a \in M$.
By assumption, the cyclic submodule generated by $a$, denoted $(a)$, is either $0$ or $M$.
Since $a \neq 0$ it must be that $(a) = M$, so $M$ is cyclic.
Let $f \in \mathrm{End}_R(M)$.
Since $\mathrm{ker}(f)$ and $\mathrm{im}(f)$ are submodules of $M$,
by assumption they can be either $0$ or $M$.
If $f$ is non-zero, then $\ker(f) = 0$ and $\mathrm{im}(f) = M$,
so $f$ is bijective, and hence has an inverse.
Therefore, $\mathrm{End}_R(M)$ is a division ring.

Addressing the converse, consider $\mathbb{Q}$ as a $\mathbb{Z}$ module. Any rational number generates a nontrivial submodule, so $\mathbb{Q}$ is not simple. But by what follows, we can see that $\mathrm{End}_\mathbb{Z}(\mathbb{Q}) \simeq \mathbb{Q}$. Take some $\phi \in \mathrm{End}_\mathbb{Z}(\mathbb{Q})$. Since \begin{equation*} \phi\left(\frac{a}{b}\right) = a\phi\left(\frac{1}{b}\right) \end{equation*} we only have to define $\phi$ on rational numbers with a numerator of $1$. But also \begin{gather*} \phi\left(1\right) = \phi\left(\frac{b}{b}\right) = b\phi\left(\frac{1}{b}\right) \\ \Longrightarrow \quad \frac{1}{b} \phi\left(1\right) = \phi\left(\frac{1}{b}\right) \,, \end{gather*} so really we only have to specify $\phi(1)$, which can be any rational number.