The answer is yes, we can, and, in fact, they are precisely bounded distributive lattices.
Lemma 1: FOIL = FIOL
Middle schoolers are often taught the acronym "FOIL" to remember how to distribute binomials over each other: "First, Outer, Inner, Last" refers to the order in which we take pairs from each binomial's terms. It turns out that even when neither operation is commutative, "FIOL" is still a perfectly valid way to distribute the binomial! Observe:
$$(x \times_1 y) \times_2 (z \times_1 w) \\
= (x \times_2 (z \times_1 w)) \times_1 (y \times_2 (z \times_1 w)) \\
= (x \times_2 z) \times_1 (x \times_2 w) \times_1 (y \times_2 z) \times_1 (y \times_2 w) \quad \text{(FOIL)}$$
$$(x \times_1 y) \times_2 (z \times_1 w) \\
= ((x \times_1 y) \times_2 z) \times_1 ((x \times_1 y) \times_2 w) \\
= (x \times_2 z) \times_1 (y \times_2 z) \times_1 (x \times_2 w) \times_1 (y \times_2 w) \quad \text{(FIOL)}
$$
This gives FOIL = FIOL, as desired.
Lemma 2: M obeys the absorption laws and both operations are idempotent
This is straightforward to show, but requires some cleverness with the neutral elements:
$$x = x \times_1 e_1 \\
= x \times_1 (e_1 \times_2 y) \\
= (x \times_1 e_1) \times_2 (x \times_1 y) \\
= x \times_2 (x \times_1 y)
$$
Because each neutral element is a double-sided identity/zero, we can change which side we place the $e_1$ and/or $e_2$ on to arbitrarily commute this identity. (This is important for later!) We can also easily repeat this argument with swapped subscripts on the multiplication operations and neutral elements to yield the dual absorption law. Finally, note that we can substitute $y = e_1$ (resp. $e_2$) in the identity to yield $x = x \times_2 x$ (resp. $x = x \times_1 x$) to get idempotency, as desired.
Final step: commutativity
The final trick to conclude this proof is applying FOIL = FIOL to
$$(x \times_1 y) \times_2 (y \times_1 x)$$
to yield both
$$(x \times_2 y) \times_1 x \times_1 y \times_1 (y \times_2 x) \qquad \text{(FOIL)}$$
and
$$(x \times_2 y) \times_1 y \times_1 x \times_1 (y \times_2 x) \qquad \text{(FIOL)}$$
and after applying absorption, we yield $x \times_1 y = y \times_1 x$, as desired. (The dual argument is obvious, and now we clearly have the axioms of a bounded distributive lattice.)
