Does this "reverse distributivity" ever occur: $a \circ (b\times c) = (a \times b) \circ (a \times c)$?

I will treat the case where $\times$ forms a monoid with unit element $1$. From monoids, one can go in two directions: towards group-like structures or towards lattice-like structures. The group-like cases are all trivial, but we can find many non-degenerate lattice-like examples.

So consider a monoid $(M,\times,1)$ and assume that we have an operation $\circ$ that satisfies the reverse distributivity law $a\circ(b \times c) = (a \times b)\circ(a \times c)$ for any $a,b,c \in M$. It follows then that have $b\circ c = (1 \times b)\circ(1\times c)=1\circ(b\times c)$ for all $b,c \in M$.

Define the function $f: M \rightarrow M$ as $f(x) = 1\circ x$. Then reverse distributivity becomes $f(a \times b \times c) = a\circ (b \times c) = (a \times b) \circ (a \times c) = f(a \times b \times a \times c)$. Vice versa, any function $f: M \rightarrow M$ satisfying $f(a \times b \times c) = f(a\times b \times a \times c)$ for all $a,b,c \in M$ gives rise to a reverse distributive operation via $a\circ b = f(a \times b)$. Therefore:

1. Any potential $f$ must satisfy $f(a) = f(a \times a)$ for all $a \in M$. This follows by setting $b,c = 1$ in the equation above.

2. If $M$ is a group, then all such functions $f$ are in fact constant. To see this, notice that $f(a\times b \times c) = f(a \times b \times a \times c)$, then set $c=1$ and $b=a^{-1}$ to conclude $f(1) = f(a \times a^{-1}) = f(a \times a^{-1} \times a) = f(a)$ for all $a \in M$.

3. If $M$ is an idempotent commutative monoid (semilattice), then the equality $f(a \times b \times c) = f(a \times a \times b \times c) = f(a \times b \times a \times c)$ holds for any function $f: M \rightarrow M$. This observation yields plenty of non-degenerate examples. For example, consider the Boolean algebra $\{\text{true},\text{false}\}$ of truth values, and set $\times$ to be disjunction $\vee$. Define $a \not\vee b = \neg(a \vee b)$. The pair $(\vee,\not\vee)$ satisfies reverse distributivity, i.e. $a \not\vee (b \vee c) = (a \vee b) \not\vee (a \vee c)$. Similarly for $\wedge$.

If the structure $(M,\times)$ is a general semigroup, then these observations fail. Interestingly, by analyzing all 14 non-constant binary operations, we can show that all suitable pairs of operations $(\times, \circ)$ over truth values have $a \circ b = f(a \times b)$ for some $f$. However, there is a three-element structure $(S,\times,\circ)$ with $(\times,\circ)$ satisfying reverse distributivity that does not have $a \circ b = f(a \times b)$ for any function $f$ and with $a\times b$ non-constant. E.g. Consider the two operations below.

x|012    o|012
-----    -----
0|121    0|011
1|211    1|011
2|111    2|011

Since $b \times c \neq 0$, we have $a \circ (b \times c) = 1 = 1 \circ 1 = (a \times b) \circ (a \times c)$, but not $a \circ b = f(a \times b)$, since that would give $0 = 0 \circ 0 = f(0 \times 0) = f(1) = f(1 \times 1) = 1 \circ 1 = 1$.

edit: While I'm sure that someone, somewhere has already named these structures, I doubt that there's a widely-agreed-upon name for them, even among people who actively study these structures (especially given that we don't have such a name even for the self-distributive case, and even when equipped with additional axioms. Viz. racks, crystals, automorphic sets, etc.).