Are algebraic structures uniquely identifed by their free objects? It might be a naive question, as I am not a specialist in this field.
This is a follow-up to this question.
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?
 A: I think you seems to be interested in quasivarieties or quasipseudovarieties.  
Let me stick to the non-pseudo world. For the pseudo-setting you need to always use profinite analogs. 
While a variety is determined by its free object on a countable generating set (assuming all operations have finite arity) quasivarieties are highly non determined by free objects. For instance the cancellative laws are quasiidentities and so cancellative monoids form a quasivariety. But the free cancellative monoid is the same as the free monoid so the free object in the much larger quasivariety of all monoids is already free in the much smaller quasivariety if cancellative monoids. 
A: Two non-isomorphic algebras $A$ and $B$ may generate the same variety and so the free objects of the variety generated by an algebra $A$ do not characterize $A$ uniquely. For example, for any group $A$, we have $Var(A)=Var(A\times A)$, so free objects are the same however $A$ and $A\times A$ are not isomorphic. In general the classes of algebras like the one you are dealing with are quasi-varieties and so they may not contain free objects (for example, note that there is no free field, since the class of fields is not a variety, it is just a quasi-variety). But it seems that the special class you introduced is a pre-variety too (it is closed under subalgebra and product) so it contains free objects (you should check it out if the class is really closed under product or not). For more details see Burris, Sankapanavar: A course in universal algebra.
