Suppose $G$ is a finitely generated group, $A$ is a finite set of generators of $G$. Define $\pi: (A \cup A^{-1})^* \to G$ using the following recurrence:
$$\pi(\Lambda) = e$$ $$\pi(a \alpha) = a\pi(\alpha)$$
Now define the language $L(G, A) := \{w \in (A \cup A^{-1})^*| \pi(w) = e\}$.
Note that the position of $L(G, A)$ (as well as of $(A \cup A^{-1})^* \setminus L(G, A)$) in Chomsky hierarchy is uniquely determined by the properties of $G$.
It is a well known fact, proved by Novikov and Boone, that there exist such groups $G$, that $L(G, A)$ is recursively enumerable, but $(A \cup A^{-1})^* \setminus L(G, A)$ isn't. Such groups can even be finitely presented. The smallest known such group has $12$ generators.
My question is:
Do there exist such groups $G$, such that $(A \cup A^{-1})^* \setminus L(G, A)$ is recursively enumerable, but $L(G, A)$ isn't?
I know, that $L(G, A)$ its recursively enumerable iff $G$ is isomorphic to a subgroup of a finitely presented group.
I also know that a such asymmetry holds for context-freedom: on one hand context free $L(G, A)$ implies context free $(A \cup A^{-1})^* \setminus L(G, A)$, but on the other hand, there exist groups such that $(A \cup A^{-1})^* \setminus L(G, A)$ is context free, but $L(G, A)$ isn't.