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Is the variety of algebras $(A,*,+)$ that satisfy $(x*y)+(y*z)=(x+y)*(y+z)$ generated by its finite algebras?

The finite algebras $(A,*,+)$ that satisfy the identity $(x*y)+(y*z)=(x+y)*(y+z)$ are precisely the algebras such that the one-dimensional cellular automata produced by $*$ and $+$ are commutative cellular automata. To be clear the operations $*,+$ do not necessarily satisfy any associativity or any other well known laws. See the paper Commuting Cellular Automata for details on this identity and its relation to cellular automata (or see my answer here for less details). I want to know if the variety $V$ of algebras that satisfy $(x*y)+(y*z)=(x+y)*(y+z)$ is generated by the collection of all its finite algebras. It suffices to show that the free algebras in $V$ are subdirect products of finite algebras in $V$. If the finite algebras $(A,*,+)$ satisfy other kinds of identities not implies by $(x*y)+(y*z)=(x+y)*(y+z)$, then what is an explicit example of such an identity? If the free algebras in $V$ are subdirect products of finite algebras in $V$, then what is a good explicit example of how to embed the free algebras in $V$ as subdirect products of finite algebras in $V$?

Part of the motivation for this question is that cellular automata are usually over a finite alphabet $A$, so I wonder if finite algebras that satisfy $(A,*,+)$ that satisfy the identity $(x*y)+(y*z)=(x+y)*(y+z)$ satisfy any other identities.