As you noticed a solvable, simple group is necessarily abelian, since the derived subgroup is trivial. Now, let $a$ be any nonidentity element. The subgroup generated by $\{a\}$ is a normal subgroup, equal to our entire group by the hypothesis that it is simple.
If it were an infinite cyclic group (isomorphic to $\Bbb{Z}$), then there are certainly nontrivial proper subgroups, e.g. the subgroup generated by $a^2$.
Otherwise, it is isomorphic to $\Bbb{Z}_n$, and for each nontrivial divisor of $n$ corresponds a nontrivial proper subgroup. Therefore, $a$ has prime order.
It follows that the given group is prime cyclic.