Mapping reducibility from recursive to recursively enumerable language
If $L_1 \le_m L_2$, that is, $L_1$ is many-one (i.e., mapping) reducible to $L_2$ and $L_2$ is recursively enumerable (r.e.), then $L_1$ must also be r.e. This is because an acceptor for $L_1$ can be derived by reducing any instance $w$ of $L_1$ to an instance $w'$ of $L_2$ and then simulating the acceptor for $L_2$ (which exists on grounds of $L_2$ being r.e.) on $w'$.

However, in general, $L_1$ does not have to be recursive (which is strictly stronger than r.e.). In fact, any language $L$ is (trivially) many-one reducible to itself, so just take your favorite r.e. but not recursive language $L$ and note that $L \le_m L$ holds despite $L$ not being recursive.



Honestly, I believe your confusion arises from your unnecessarily complex view of many-one reducibility (i.e, on the basis of $L_1$, $L_2$, and their complements). Note that you can think of a reduction of $L_1$ to $L_2$ instead as mapping any instance $w$ of $L_1$ to an instance $w'$ of $L_2$ under the condition $w \in L_1 \iff w' \in L_2$. Stating this based on the equivalence sign "$\iff$" avoids having to think about the complements of $L_1$ and $L_2$, thus avoiding unnecessary "clutter" in your thought process.