Consider two classes of algebras(not necessarily of the same structure), say $\mathbf{A}$ and $\mathbf{B}$ and suppose the following claim:
An algebra is an $\mathbf{A}$-algebra if and only if it is a subdirect product of $\mathbf{B}$-algebras.
If $\mathbf{B}$ is closed under products and subalgebras, is it then fair to say that $\mathbf{A}$ is embedded in $\mathbf{B}$?
Well, under your assumptions, $\mathbf{A}$ and $\mathbf{B}$ are equal. Every $\mathbf{A}$-algebra is a subalgebra of a direct product of $\mathbf{B}$-algebras, and hence a $\mathbf{B}$-algebra. Conversely, every $\mathbf{B}$-algebra is a subdirect product of $\mathbf{B}$-algebras (just take the $1$-fold product consisting of the algebra itself), and thus an $\mathbf{A}$-algebra.