UPDATE As it was shown by Pasha Zusmanovich below considering only $\Box$ leads us to a trivial variety(of course we can try to proceed to a quasivariety..).
But, if we add a transposition to the signature situation becomes much more interesting.First of all we have $(A\Box B)^T=A\Box B^T$ and the left unit $I\Box A= A$Really, if we consider $T$-invariant subalgebras of some matrix algebra with $\circ$, than we can note that such algebras could be decomposed (as vector spaces) to the direct sum of Jordan algebra and Lie algebra -- symmetric and antisymmetric part,respectively.Axiomatizing this decomposition we get...Commutativity for symmetric part:(A+A^T)\Box (B+B^T)=(B+B^T)\Box (A+A^T),Power-associativity for symmetric part:(A+A^T)\Box ((A+A^T)\Box (A+A^T))= ((A+A^T)\Box (A+A^T))\Box (A+A^T)Jordan identity for symmetric part:((A+A^T)\Box (B+B^T))\Box ((A+A^T)\Box (A+A^T))= (A+A^T)\Box ((B+B^T)\Box ((A+A^T)\Box (A+A^T)))Anticommutativity for antisymmetric part:A\Box A+A^T\Box A^T=A\Box A^T+A^T\Box ALie identity for antisymmetric part:(A-A^T)\Box ((B-B^T)\Box (C-C^T))+(C-C^T)\Box ((A-A^T)\Box (B-B^T))+(B-B^T)\Box ((C-C^T)\Box (A-A^T))=0$$
Commutativity and power-associavity for symmetric part could be seen as averaged commutativity and averaged associativity and(!) commutativity, respectively.A\Box B + A\Box B^T + A^T\Box B + A^T\Box B^T= B\Box A + B^T\Box A + B\Box A^T +B^T\Box A^T\sum_{\sigma\in S_3}\sum_{(i,j,k)\in (\varnothing,T)^3}A_{\sigma(1)}^{i}\Box(A_{\sigma(2)}^j\Box A_{\sigma(3)}^k)=\sum_{\sigma\in S_3}\sum_{(i,j,k)\in (\varnothing,T)^3}(A_{\sigma(1)}^{i}\Box A_{\sigma(2)}^j)\Box A_{\sigma(3)}^k$$
Did anyone consider something close to varieties of algebras with identities of that type?
