Structures with $x*(y*z) = y*(x*z)$

This axiom is rather weak on its own and I don't think there is very much that is interesting which you can say about it. There is a very important qualitative difference between this axiom and other more familiar axioms like commutativity or associativity which have a rich theory, which is that in the equation $$x*(y*z) = y*(x*z)$$ each variable either only appears as the left input to $*$ or only appears as the right input, and moreover outputs only appear as right inputs. As a result, you can pretend that the left inputs and right inputs are separate "types" which just happen to be labelled by the same set, with the operation $*$ just giving a way for the left inputs to act as functions on the right inputs.

To be more precise, such a magma $(X,*)$ is really equivalent to giving a collection $Y$ of commuting functions $X\to X$ together with a (surjective) function $f:X\to Y$. Indeed, given such data, you can define $x*y=f(x)(y)$; conversely, given such a magma, you can let $Y$ be the set of all functions of the form $y\mapsto x*y$ and let $f$ be the map sending $x$ to $y\mapsto x*y$.

(In particular, if $(X,*)$ is such a magma and $f:X\to X$ is any function, then $(X,\cdot)$ is also such a magma where $x\cdot y=f(x)*y$. This gives a huge wealth of examples.)

So all you really have is some collection of commuting maps from $X$ to itself, together with a (completely arbitrary!) way to assign such maps to elements of $X$. I can't imagine any interesting theory coming from the latter arbitrary assignment, so you might as well just be studying collections of commuting maps on a set.

One way to define the (Cartesian) product of two objects in a category is as the adjoint to the $Hom$ functor; That is, Cartesian products are intended to obtain a bijection $Hom(A \times B,C)\simeq Hom(A,Hom(B,C))$. Since $A \times B $ is isomorphic to $B \times A$, the above isomorphism implies, in turn, a bijection between $Hom(A,Hom(B,C))$ and $Hom(B,Hom(A,C))$. Thus the operation $Hom$, modulo bijection, satisfy your algebraic property.