I think the answer is no. Fix a nonabelian finite simple group $S$ and a sequence $(m_n)$ of integers at least 2. Define inductively $G_1=S$ and $G_{n}=S^{m_n}\wr G_{n-1}$. This group admits only $S=G_1$ as simple quotient and only $S$ as Jordan-Hölder factor. I claim that, provided $(m_n)$ grows fast enough, the rank (minimal number of generators) of $G_n$ is unbounded, so that the inverse limit of the $G_n$ is not finitely generated, although it satisfies your assumptions.
The claim that $G_n$ has a unique maximal normal subgroup $W_n$ is obtained by an easy induction. This uses only that $G_n=V_n\rtimes G_{n-1}$, where the only subgroups of $V_n$ that are normal in $G_n$ are $V_n$ and {1} (we need $S$ be nonabelian here) and $V_n$ has trivial centralizer in $G_n$.
On the other hand, we use that the rank $r(G)$ of a wreath product $G=A\wr B$ is at least $r(A)/|B|$. Indeed, Suppose that $G$ is generated by $r$ elements. Then it's generated by $B$ and $r$ elements of $A^B$, $a_i=(a_{ij})_{1\le j\le |B|}, i=1\dots r$. So $A$ is generated by the $a_{ij}$. Thus $r(A)\le |B|r(G)$.
Since the rank of $S^n$ tends to infinity when $n$ tends to infinity, we can define inductively the sequence $(m_n)$ so that the rank of $G_n$ is at least $n$.
Edit: as pointed out by Maurizio in an email, in the example it's not true that the only subgroups of $V_n=S^{m_n}$ that are normal in $G_n$, are $V_n$ and the trivial group. Actually there are a little more: $T^{m_n}$, where $T$ is a direct factor in $S^{m_n}$. This is however enough to run an induction that $G_n$ has a unique maximal normal subgroup.
