Properties of finite magmas $(S,\cdot)$ with $\forall(x,y,z)\in S^3, x\cdot(y\cdot z)=y\cdot(x\cdot z)$? I am considering finite magmas $(S,\cdot)$ with $\forall(x,y,z)\in S^3, x\cdot(y\cdot z)=y\cdot(x\cdot z)$. Any finite commutative group is an example of such thing. But in the context (this question on crypto.stackexchange.com), I am not interested in groups; or at least, not in groups with an efficiently computable inverse.
I am wondering if/how the classical magma-to-group classification simplifies for such structures.

 A: It is related to commutativity: if there's an identity element, then it implies it (by putting $z=e$), if we have associativity, then it is implied by it (obviously). On the other hand, the property along with commutativity implies associativity (it is easy to check), so a loop with this property is already a commutative group, and any magma with identity and  this property is automatically a commutative monoid.
Any finite set with operation of right projection $a\cdot b=b$ has the property and is associative, but not divisible and does not have identity (if the set has at least 2 elements), so that takes care of two arrows. (It is also not commutative.)
A finite linear order with the operation $a\vee b=\max(a,b)$ is a noninvertible monoid with the property, which takes care of the lower right one.
As per Generic Human's answer, a simple example of a quasigroup without identity (albeit with a left identity) that is nonassociative, but having the property in question, is the cyclic group $\mathbf Z_3$ with the operation $x\cdot y=y-x$ (or any commutative group with an element of order at least 3 with the same operation), so that is the complete picture.
I'm opening a community wiki, so others can contribute.
A: About quasigroups
The submagma $M_S$ of $S\to S$ generated by left multiplication $\phi_x: y\mapsto x\cdot y$ is commutative. If $S$ is a finite quasigroup, $M_S$ is only composed of permutations (as $\phi_x(x\backslash y)=y$ for all $y$), so it is a (commutative) subgroup of the symmetric group $\mathcal S_n$. In particular for each $x$ there is an $x^{-1}$ such that $x^{-1}\cdot(x\cdot y)=y$.
Every commutative subgroup $H$ of $\mathcal S_n$ ($|H|\ge 3$) can be associated to a quasigroup without identity in this way: pick a permutation $\pi$ of the $|H|$ elements of $H$ distinct from the identity such that $\pi(1)=1$. Then the operation $x\cdot y=\pi^{-1}(x\circ\pi(y))$ defines a quasigroup structure, with $x\backslash y=x^{-1}\cdot y$ and $x/y=\pi(x)\circ\pi(y)^{-1}$. Obviously $1\cdot x=x\ne x\cdot 1$.
In fact this is exactly the set of quasigroups $Q$ without identity such that $x\mapsto \phi_x$ is injective, since necessarily $Q=M_Q$ and $\pi(x)=x\cdot 1$.
If $H$ has an element of order greater than 2, we can choose $\pi(x)=x^{-1}$ and $\cdot$ boils down to left division in $H$: $x\cdot y=x^{-1}\circ y$. So $C_3$ equipped with left division is a simple example of a quasigroup satisfying the condition.
