$\mathrm{P\Sigma L}(2,8)$ is a counter-example.

A corrected version of the statement is that, under the hypothesis, either

a) $G\cong Z_p$ for some prime $p$,

b) $G\cong Z_p\times Z_q$ for some primes $p$ and $q$,

c) $G$ is a Frobenius group with kernel an elementary abelian $p$-group and the complement has prime order, or

d) $G$ is an almost simple group.

Proof:

Let $N$ be a minimal normal subgroup of $G$. We have $N=T_1\times\cdots \times T_k$, where the $T_i$'s are isomorphic simple groups. Take $L\leq T_1$ and let $H$ be as given by the hypothesis.

Case 1) $H\leq N$. In this case, we have $G=N$. If $T$ is abelian, then $G$ is abelian hence $k=1$ and we are in case a). If $T$ is nonabelian then, since $N$ is generated by $T_1$ and $H$ a group of prime order, we find that $k=1$ and $G$ is simple.

Case 2) $H\not\leq N$. Since $H$ has prime order, we have $N\cap H=1$ and thus $G=N\rtimes H$. If $T$ is abelian, then either $H$ is malnormal and $G$ is Frobenius and we have c), or $G$ is abelian and we have b). If $T$ is nonabelian, then the $T_i$'s are the only minimal normal subgroups of $N$ and $H$ must act transitively on $\{T_1,\ldots,T_k\}$. As $H$ has prime order, either $k=1$ (and we are in case d)), or $k=|H|$. In the latter case, $L$ has exactly $k$ conjugates, each of them contained in some $T_i$. On the other hand, these conjugates must generate $N$, which is a contradiction.

Note that this not exactly a characterisation. For example, not all almost simple groups have this property and neither do all groups in c).