It might be a naive question, as I am not a specialist in this field.
This is a follow-up to [this question](https://mathoverflow.net/questions/137640/shape-of-axioms-in-abstract-algebra).

I want to study varieties of objects generalizing ordered monoids, in particular using an explicit operator $\omega$, subject to the axiom $(A)$ $xx=x\implies x^\omega=x$, i.e. the restriction of $\omega$ to idempotents has to be the identity function.

Does it make sense to talk about a free algebra in such a context? The problem is that this free algebra would be the same as the one where we don't specify axiom $(A)$, and replace it by axioms $(A')$ like $x^\omega=x^\omega x^\omega$ and $(x^\omega)^\omega=x^\omega$.

If two algebraic structures have exactly the same free object, it seems like this object does not really represent the constraints given by the axioms, and therefore the axioms are not of a shape that allows to define a free object. Is this reasoning valid? 

For now, the approach taken is to use axioms $(A')$, and to later use a profinite equation $x^\omega=x^\pi$, where $\pi$ is the profinite idempotent power, in order to define the pseudovariety of objects we are really interested in. But I'm wondering if such a detour is necessary.

**EDIT**

More detailed question:
If we were to use axiom (A), what is needed to be proved in order to show that there is a free object? It seems that closure under arbitrary product is important, why is it the case?
Also, if we just want the universality property for finite objects, then is it enough to prove closure under finite products? Also, isn't enough to take as free object the set of terms quotiented by the axioms, and directly show that it has the universal property?