Just wanted some feedback to ensure I did not make any mistakes with this proof. Thanks!
Since $G$ is abelian, every subgroup is normal. Since $G$ is simple, the only subgroups of $G$ are $1$ and $G$, and $|G| > 1$, so for some $x\in G$ we have $x\neq 1$ and $\langle x\rangle\leq G$, so $\langle x\rangle = G$. Suppose $x$ has either infinite order or even order. Then $\langle x^2\rangle \leq G$ but $\langle x^2\rangle \neq \langle x \rangle$, a contradiction. So $x$, and therefore $G$, has odd order, and by Feit-Thompson, $G\cong Z_p$ for some prime $p$.
Edit: Thanks, I see that Feit-Thompson is too much.
Since $G$ is abelian, every subgroup is normal. Since $G$ is simple, the only subgroups of $G$ are $1$ and $G$, and $|G| > 1$, so for some $x\in G$ we have $x\neq 1$ and $\langle x\rangle\leq G$, so $\langle x\rangle = G$. Suppose $x$ has infinite order. Then $\langle x^2\rangle \leq G$ but $\langle x^2\rangle \neq \langle x \rangle$, a contradiction. So $x$, and therefore $G$, has finite order. Suppose $x$ has composite order $n$ so for some $p > 1$ that divides $n$, $\langle x^p \rangle$ is a proper non-trivial subgroup of $G$, so $G$ is not simple. So $G$ is a cyclic group of prime order.
Best Answer
Er okay: suppose $G$ is simple, abelian. Then any $g \in G$ has to generate the whole thing, i.e. we have $\mathbb{Z} \twoheadrightarrow G$. $\mathbb{Z}$ is not simple (consider say $2\mathbb{Z}$), so $G \simeq \mathbb{Z}/n \mathbb{Z}$ for some $n$. If $n$ is not prime, take any $p \mid n$, we have $\langle p\rangle \subseteq \mathbb{Z}/n\mathbb{Z}$ is a proper subgroup.