If L is regular, so is L2 (proof using closure properties)
Here are several possible ways of proving that if $L$ is regular then so is $M$.

Method 1

Show first that the language
$K = \{ a^r : |s|_a=r \text{ for some } s \in M \}$.
You can obtain $K$ from $L$ by deleting all letters other than $a$. Then show that $M$ is regular by "undeleting" all letters other than $a$.

Method 2

Construct an NFA for $M$ directly, given an NFA for $L$. At each step, the new NFA accepts non-$a$ letters gratuitously, both in the input and in its simulation of the NFA for $L$. It also reads $a$, but then must also simulate an $a$ transition on the NFA for $L$. Finally, it accepts whenever it can get the NFA for $L$ to be at an accepting state.

Method 3

Parikh's theorem shows that $K$ (defined in Method 1) is eventually periodic. That is, for some integers $m,p$ and sets $A \subseteq \{0,\ldots,m\}$, $B \subseteq \{0,\ldots,r-1\}$ we have
$$
K = \{ a^i : i \in A \} \cup \{ a^{m+rj+i} : j \geq 0, i \in B \}.
$$
Given this, we can construct a DFA for $M$ directly (exercise).