First of all, please avoid the term $\textrm{co-}A_{TM}$, it is confusing (co-X means the *class* of languages complementary to languages in the *class* X). Better to write A_{TM}-complement.

Also, the notation $\textrm{co-}(\mathcal{RE} \cup \textrm{co-}\mathcal{RE})$ is plainly wrong, since $\textrm{co-}(\mathcal{RE} \cup \textrm{co-}\mathcal{RE}) = \mathcal{RE} \cup \textrm{co-}\mathcal{RE}$.

Now, for your question.

The solution says that there is an invertible mapping reduction between $L$ and $A_{TM}$ —- simply replace "yes" with "no" and vice versa.

The fact that the definition of $L$ uses $\overline{A_{TM}}$ has nothing to do with $L$ being or not in $\textrm{co-}\mathcal{RE}$; take, for instance, $L' = \overline{A_{TM}} \cap \{w: |w| < 100\}$, which is finite and thus recursive.