What equational properties of a group only need to be checked on a generating set? Let $G$ be a group and $S\subset G$ a generating set. Let $P$ (short for $P(x_1,\dots,x_n) = 1$) be an equational property that may or may not be satisfied by all $n$-tuples of elements of $G$. My question is:

What $P$ are such that $P$ being satisfied on $S$ implies that it is satisfied on all of $G$? 

Is there a simple characterization? (If so, do you know a proof?) Is this a hard question with a developed theory? (If so, can you point me to relevant sources?)
Examples of $P$'s: 
(1) If $P$ is $[x_1,x_2]= x_1x_2x_1^{-1}x_2^{-1}=1$, then $P$ being satisfied on $S$ implies that it is satisfied on $G$. Indeed if pairs of generators commute, then words on those generators clearly commute since the generators can "slip past each other."
(2) On the other hand, if $P$ is $x_1^2=1$, then $P$ being satisfied on $S$ says nothing about $G$. For an extremely hands-on example, $S_3$ is generated by two elements of order $2$, but not all its elements are order $2$.
Motivating example: 
What got me interested in this in the first place was that I was wondering if free metabelian groups (on a finite number of generators) are finitely presented. Does the relation $[[x_1,x_2],[x_3,x_4]] = 1$ that defines metabelianness carry over from a generating set to the whole group? A few google searches seems to have revealed that it does not, so now I'm wondering what the other relations might be that do carry over.
Thanks in advance.
 A: This is more of an extended comment:
If your group has two generators, like the free group on two generators, then these metabelian relations hold for the generators, since $[[x,y],[x,y]]=1,[[x,y],[y,x]]=1,[[x,y],[x,x]]=1$ (this is just asking if $[x,y]$ commutes with itself, with its inverse, and with the identity). 
You should look into verbal subgroups and reduced free groups, I know that Combinatorial Group Theory by Karrass, Magnus, Solitar has some stuff on them. It is known that reduced free groups all have presentations $\langle a_1,...,a_n \mid X_1^d, P_1,....,P_i,...\rangle$ where the $X_1,P_i$ are all the words that satisfy the equation they specify (so $P=X_1X_2X_1^{-1}X_2^{-1}$ is the commutator subgroup and $X_1^d$ is the group generated by all $d$-powers), and $P_i$  are in the commutator subgroup (this is Cor 2.3.1 in Combinatorial Group Theory). You can find by looking directly at any original presentation and transforming the presentation in a fairly simple manner (although I am not sure if "defining relation on generators is enough"-property is preserved, but it seems plausible).
I wouldn't be surprised if the commutator equation is the only one where you can get away with generator only relations, and feel like there should be a way to finagle an answer using that corollary.
Update 1: The equation for the equation $[X,Y]X^p=1$ it is enough to define on generators (where $X,Y$ vary over the group.) This is because if you have generators $x_1,...,x_n$, you can look at $[x_i,x_i]x_i^p=1=x_i^p$, so your generators are torsion, and $[x_i,x_j]x_i^p=1=[x_i,x_j]$ (since $x_i^p=1$) so your group is commutative. So your group is $\langle x_1,...,x_n \mid x_i^p, [x_j,x_k]; 1\leq i,j,k \leq n \rangle $ which is exactly the free reduced group version (ranging over all elements rather than just the generators).
Note that if you change this to something like $[X,Y]W(X,Y)=1$ ,where $W(X,Y)$ is a word in $X,Y$ you still get that $x_i$ are torsion provided $W(X,X)$ is not in the commutator subgroup, by the same argument, but you don't know, a priori, that $W(x_i,x_j)$ will be trivial, so you won't be able to conclude $[x_i,x_j]=1$, which is important to the argument. The reduced free group version of this group will be torsion( provided $W(X,X)$ is not in the commutator group) using the techniques in the proof of corollay 2.3.1. It feels very unlikely if you got a group which ranged over just the generators you would end up with a torsion group for most choices of $W(X,Y)$.
