Any family of language that is a trio is
closed under interleaving with a regular set.
This includes of course interleaving of 2 regular sets, since regular sets form a trio.
Proving the result (and more) only with closure properties
Note: I created the definitions below for the purpose of this
question. I do not know whether there are established definitions for
this, which might exist under another name.
The purpose of this approach is to avoid any complex construction of
automaton. But we need at least one specific operation to account for
dealing with several strings at the same time. And, as an unexpected benefit, the end result is
much more general (this is actually more to be expected from proofs based on closure properties). However, the proof is centered on the question asked, and only includes remarks to show how it generalizes.
Consider two alphabets $\Sigma_i$ for $i=1,2$. We can consider their
product $\Sigma_1\times\Sigma_2=\{(a_1,a_2)\mid a_1\in\Sigma_1\wedge
a_2\in\Sigma_2\}$ as a new alphabet, where the symbols are pairs of
symbols of $\Sigma_1$ and $\Sigma_2$.
Similarly, with 3 alphabets, we can build an alphabet of triples
(instead of pairs). We ignore the trivial issue of associativity in
using pairs to build triples, or $n$-tuples, here and in the rest of
this answer.
Now, given two strings $x\in\Sigma^*$ and $y\in\Pi^*$ such that $|x|=|y|$
we can define the conflation of these two strings as the string $z=\mathrm{Conflate}\,(x,y)\in(\Sigma\times\Pi)^*$
with the same size, such that $\forall i\in[1,|x|], z_i=(x_i,y_i)$.
We can similarly conflate $n$ strings of equal length into a single
string of $n$-tuples of symbols ... but we will not go beyond $n=3$.
Finally, given two languages $L_1\subseteq\Sigma_1^*$ and $L_2\subseteq\Sigma_2^*$
we can define the conflation of these two languages:
$$\mathrm{Conflate}\,(L_1,L_2)=\{\mathrm{Conflate}\,(x,y)\mid
|x|=|y|\wedge x\in L_1 \wedge y\in L_2\}$$
We can also conflate similarly any number of languages, to produce a
language on the cross product of their alphabets.
This $\mathrm{Conflate}$ operation has many simple properties, that
are rather trivial to prove.
Given two alphabets $\Sigma_1$ and $\Sigma_2$ and two languages
$L_1\subseteq\Sigma_1^*$ and $L_2\subseteq\Sigma_2^*$:
$\mathrm{Conflate}\,(L_1,L_2)\subseteq(\Sigma_1\times\Sigma_2)^*$
$\mathrm{Conflate}\,(L_1,\Sigma_2^*)$ is regular iff $L_1$ is regular
$\mathrm{Conflate}\,(\Sigma_1^*,L_2)$ is regular iff $L_2$ is
regular
The proof uses a projection homomorphism that keep only the $L_1$ or the $L_2$ component of the conflation.
side note: the above is also true for context-free, and more
generally families of languages
closed under non-erasing homomorphism and inverse homomorphism (such as trios). For example, if $\mathcal F$ is a trio, and $L$ is a language, and $\Sigma$ and alphabet (not necessarily the alphabet of $L$), then $\mathrm{Conflate}\,(L,\Sigma^*)\in\mathcal F\;$ iff $\;L\in\mathcal F$.
$\mathrm{Conflate}\,(L_1,L_2)= \mathrm{Conflate}\,(L_1,\Sigma_2^*)
\cap \mathrm{Conflate}\,(\Sigma_1^*,L_2)$
Hence, if $L_1$ and $L_2$ are both regular, then
$\mathrm{Conflate}\,(L_1,L_2)$ is also regular.
Now we consider also the alphabet $B=\{0,1\}$, and the alphabet
cross-product $\Sigma_1\times\Sigma_2\times B$, and we define on this
alphabet the substitution $\sigma$ as follows:
$\forall (a_1,a_2,b)\in(\Sigma_1\times\Sigma_2\times B),\;
\sigma((a_1,a_2,b))=\;($ if $b=0$ then $a_1$ else $a_2)$.
If $L_1$ and $L_2$ are both regular, then
$\mathrm{Conflate}\,(L_1,L_2)$ is also regular, and thus
$\mathrm{Conflate}\,(\mathrm{Conflate}\,(L_1,L_2),B^*)$ is regular,
since $B^*$ is.
Applying the substitution $\sigma$, since regular sets are closed
under substitution, we know that the language
$\sigma(\mathrm{Conflate}\,(\mathrm{Conflate}\,(L_1,L_2),B^*))$
is regular.
But it can fairly easily be proved that
$$\mathrm{Interleave}\,(L_1,L_2)=\sigma(\mathrm{Conflate}\,(\mathrm{Conflate}\,(L_1,L_2),B^*))$$
Hence $\mathrm{Interleave}\,(L_1,L_2)$ is regular.
The interesting point here is that, with nearly no further work, we
can prove identically that many families of languages, context-free for example,
and more generally trios (hence also AFLs), are closed under
$\mathrm{Interleave}$ composition with regular languages, notably
because the substitution $\sigma$ is non-erasing. This essentially
follows from the fact that trios are closed under inverse homomorphism, under
intersection with regular sets, and under non-erasing homomorphism.
