What can we say about generating subsets if all free subsets are finite Definitions
An (universal) algebra is a pair $\mathcal A=(A, (f_1,\dots, f_n))$ where $A$ is a non-empty set and $(f_1, \dots, f_n)$ is a family of finitary operations on $A$. The notation $o(f_i)$ will be used for the arity of $f_i$.
Given a subset $X$ of $\mathcal A$, the subalgebra of $\mathcal A$ generated by $X$, $\langle X\rangle$, is the smallest superset of $X$ stable by all $f_i$ functions. A subset $X$ is called generating iff $\langle X \rangle=A$.
A map $h:A\to B$ is called a homomorphism between $\mathcal A=(A, (f_1,\dots, f_n))$ and $\mathcal B=(B, (g_1,\dots, g_n))$ iff for all $i$ and $a_1,\dots, a_{o(f_i)} \in A$, $h(f_i(a_1,\dots, a_{o(f_i)}))=g_i(h(a_1),\dots, h(a_{o(f_i)}))$. A subset $X$ of $\mathcal A$ is called free if any function with domain $X$ can be extended to a homomorphism. (Obviously, we only consider functions that have an algebra with the same signature as codomain)

Question
What can be said about generating subsets when all free subsets are finite?
(It'd be really awesome if it implied the existence of a finite generating subset)
 A: Your definition of free subset can be reworded. Note that if $X\subseteq A$
is  free, then there is a homomorphism $h\colon A\to F(X)$
where $F(X)$ is the absolutely free algebra over $X$ in this
signature and $h$ is the identity on $X$. Conversely, if there is such a
homomorphism, then it is easy to check that $X$ is a free
subset.
Note also that
if such $h$ exists, then the freeness of $F(X)$ guarantees the existence
of a homomorphism $k\colon F(X)\to A$ that is the identity on $X$.
One can show that
(i) the map $e = kh$ is a retraction of $A$ onto the subalgebra
of $A$ generated by $X$, and
(ii) this subalgebra $\langle X\rangle$ is isomorphic to
$F(X)$; i.e. it is absolutely free over $X$.
Conversely it is not too hard to prove that if $A$ has a retraction
onto the subalgebra $\langle X\rangle$ and this subalgebra is absolutely
free over $X$, then $X$ is a free subset.
It might be easier to spot free subsets now, and to see how rare they are.
For example, if $X$ is a free subset of $A$, then $A$ can satisfy no
nontrivial identity in $|X|$ or fewer variables, since it has a subalgebra
$\langle X\rangle$ that satisfies no such identity.
This shows that, for example, no group, ring, module,
or Boolean algebra has any free subsets at all. Not even the empty
subset is free. [For groups, the identity $1\cdot 1 = 1$ holds,
and involves zero variables, so no subset of a group
containing zero or more elements can be free.] No nonempty subset of a semigroup or lattice can be free, since
these algebras satisfy nontrivial 1-variable identities. [E.g.
$x(xx)=(xx)x$, $x\wedge (x\wedge x)=(x\wedge x)\wedge x$ respectively.] 
Thus, assuming that all free subsets are finite is often
a vacuous assumption. 
To get back to the question you asked, having all free subsets finite does not imply the existence of a finite generating subset. For a counterexample, any uncountable group will not have any free subsets nor any finite generating subsets.
