**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 $\mathcal V$.