Theorem. If $\mathcal V$ is a variety of finite signature and $\mathcal V$ can be axiomatized by identities of the form $s\approx t$ where $\textit{length}(s)=\textit{length}(t)$, then $\mathcal V$ is generated by its finite members.
(For this theorem, say that $\textit{length}(s)=\textit{length}(t)$ if $s$ and $t$ have the same multiset of variables and the same number of operation symbols.)
Proof. Under the length hypothesis of the theorem, there is a congruence $\theta_n$ on ${\mathbf F}_{\mathcal V}(m)$ which relates two elements iff they are equal or else they can be each represented by terms of length greater than $n$. The signature is finite, so there are finitely many elements of ${\mathbf F}_{\mathcal V}(m)$ represented by terms of length at most $n$, and therefore ${\mathbf F}_{\mathcal V}(m)/\theta_n$ is finite. The intersection of the $\theta_n$'s is trivial, so this yields a representation $$ {\mathbf F}_{\mathcal V}(m)\hookrightarrow \prod_n {\mathbf F}_{\mathcal V}(m)/\theta_n $$ of ${\mathbf F}_{\mathcal V}(m)$ as a subdirect product of finite algebras. \\
Your variety has finite signature, and the defining identity $(x*y)+(y*z)=(x+y)*(y+z)$ has the same multiset of variables on both sides, $\{x, y, y, z\}$, and the same number ($3$) of operation symbols on both sides.
EDIT (8/18/15): The above explains why the finitely generated free algebras in $\mathcal V$ are subdirect products of finite algebras. A question was raised about whether it is necessary to also show that the infinitely generated free algebras are subdirect products of finite algebras. It is not necessary to do this: the infinitely generated free algebras in $\mathcal V$ are subdirect products of finitely generated free algebras in $\mathcal V$.