In the case $L=F$, the question can be rewritten as: let $F$ be a free group of infinite rank over the generators $(e_i)_{i\in I}$. Does there exist a characteristic subgroup $M$ of $F$ such that $F/M$ is a nontrivial finitely generated group?
The answer is no. Indeed, let $G$ be the symmetric group on the set of generators. Then the action of $G$ on the generators stabilizes $M$ and hence induces an action of $G$ on the finitely generated group $F/M$, giving rise to a homomorphism $G\to\mathrm{Aut}(F/M)$, and $F/M$ is countable because $F/M$ is finitely generated. But it is known that $G$ has no proper countable quotient (e.g., by the Schreier-Ulam theorem, see this MS post). Hence the action of $G$ on $F/M$ is trivial. This means that $e_ie_j^{-1}\in M$ for all $i,j$. Now pick two distinct elements $u,v$ in $I$, and let $\psi$ be the automorphism mapping $e_u$ to $e_ue_v$ and mapping $e_i$ to itself for any $i\neq u$. Then $\psi(e_ue_v^{-1})=e_u$. Since $e_ue_v^{-1}\in M$ and $M$ is characteristic, we deduce that $e_u\in M$; hence $e_i\in M$ for all $i$ and hence $M=F$.
