How can full theories of structures be useful in study of their equational theories?

I'm not a universal algebraist, so unfortunately I don't have any snappy applications of ultraproducts in universal algebra to share with you. But I do have a good idea of why they might be useful.

Instead of thinking about the ultraproduct as a construction that preserves sentences (since in the case of equations, the ordinary product does this just as well, as you note), think of it as an operation which preserves formulas: In an ultraproduct, you can "take the limit" of a sequence of elements from different structures, finding an element which satisfies exactly those formulas which hold only a large part of the sequence.

So for example, let's say we have an equational theory $T$ and an infinite set of equations $\{s_i(x) = t_i(x)\mid i\in \omega\}$ which don't follow from $T$ (here I'm thinking of $x$ as a singleton variable, but it could just as easily be a tuple). And let's say that for any $n\in \omega$, I can find a model $A_n\models T$ and an element $a_n\in A_n$ such that $s_i(a_n) = t_i(a_n)$ for all $i\leq n$. Then for any nonprincipal ultrafilter $U$ on $\omega$, in the ultraproduct $\prod_{n\in\omega} A_n/U$, the element $(a_n)_{n\in\omega}$ satisfies all of the equations.

The product doesn't do this for you, since an element $(b_n)_{n\in\omega}\in \prod_{n\in\omega} B_n$ only satisfies an equation if each coordinate $b_n$ satisfies the equation.

You might observe that we didn't actually need to use an ultraproduct - the reduced product $\prod_{n\in\omega} A_n/F$, where $F$ is the cofinite filter on $\omega$, does just as well. And in fact reduced products are of much more interest in universal algebra than they are in model theory. But you can imagine situations where we want to find an element in some model of $T$ which satisfies an infinite list of more complicated (first-order, but not equational) properties, and so we might need to use the full power of the ultraproduct.

Every equational theory is a first-order theory, so, at the very least, universal algebraists should be able to use the ultraproduct.

You ask why would they would need the ultraproduct when the product preserves all equations. In general, one could make the claim that universal algebraists tend to care about finite structures whereas model theorists prefer infinite structures (again, this is a generalization). The ultraproduct construction breaks down with finite structures:

If $\mathcal{K}$ is a finite set of finite algebras, then any ultraproduct of algebras from $\mathcal{K}$ is again in $\mathcal{K}$.

However, a product of algebras from $\mathcal{K}$ can be made arbitrarily large.

Two nice results that use ultraproducts in this way are a result of Jonsson, which states that a finitely-generated congruence-distributive variety is residually small, and a result of Quackenbush, which states that if a locally finite variety has only finitely many finite subdirectly irreducible algebras, then it cannot have any infinite subdirectly irreducibles.