Closure of regular languages to shuffle using closure operations
The idea is to create two copies of the alphabet $\Sigma$, say $\Sigma_1,\Sigma_2$; each $\sigma \in \Sigma$ corresponds to $\sigma^{(i)} \in \Sigma_i$. You now show that the following languages are regular:
$$
\begin{align*}
\tilde{L}_1 &amp;= \{a_1^{(1)}b_1^{(2)}\ldots a_n^{(1)}b_n^{(2)} : a_1 \ldots a_n \in L_1 \}, \\
\tilde{L}_2 &amp;= \{a_1^{(1)}b_1^{(2)}\ldots a_n^{(1)}b_n^{(2)} : b_1 \ldots b_n \in L_2 \},
\end{align*}
$$
where in both cases only some of the letters are constrained. Now take $\tilde{L}_1 \cap \tilde{L}_2$ and transform it to $L$.

I leave the rest of the details for you to complete.