[I have replaced an earlier and less complete answer]

The question depends on the precise meaning of $A_{ab}$. In the derived world, if we kill commutators then we create new potential commutators and so do not immediately end up with something commutative. Killing commutators once is the same as taking Hochschild homology. I think it is now known from work of Smith and McClure that when you apply HH to something with an action of the little $k$-cubes operad $C(k)$, you always get something with an action of $C(k+1)$ (ie it is "one step more commutative"). An associative algebra has an action of $C(1)$, so you can apply HH repeatedly and pass to a colimit to get something with an action of $C(\infty)$, which is an $E_\infty$ operad. If we are working over $\mathbb{Q}$ then $E_\infty$ algebras are essentially the same as commutative algebras. I think that this procedure converts free associative algebras to free commutative algebras.

[I have replaced an earlier and less complete answer]

The question depends on the precise meaning of $A_{ab}$. In the derived world, if we kill commutators then we create new potential commutators and so do not immediately end up with something commutative. Killing commutators once is the same as taking Hochschild homology. I think it is now known from work of Smith and McClure that when you apply HH to something with an action of the little $k$-cubes operad $C(k)$, you always get something with an action of $C(k+1)$ (ie it is "one step more commutative"). An associative algebra has an action of $C(1)$, so you can apply HH repeatedly and pass to a colimit to get something with an action of $C(\infty)$, which is an $E_\infty$ operad. If we are working over $\mathbb{Q}$ then $E_\infty$ algebras are essentially the same as commutative algebras. I think that this procedure converts free associative algebras to free commutative algebras.

UPDATE: David Ben-Zvi is right that I have misremembered results that actually apply to Hochschild cohomology (rather than homology) here. Nonetheless, what I said about repeatedly killing commutators makes some kind of intuitive sense, so I still wonder whether something along those lines could be true.

The main point here is that $A_{ab}$ is the $0$'th Hochschild homology of $A$, and that the usual derived analogue is higher Hochschild homology or topological Hochschild homology. There is a large literature about this, which is easy to find if you know the magic word 'Hochschild'.

[I have replaced an earlier and less complete answer]

The question depends on the precise meaning of $A_{ab}$. In the derived world, if we kill commutators then we create new potential commutators and so do not immediately end up with something commutative. Killing commutators once is the same as taking Hochschild homology. I think it is now known from work of Smith and McClure that when you apply HH to something with an action of the little $k$-cubes operad $C(k)$, you always get something with an action of $C(k+1)$ (ie it is "one step more commutative"). An associative algebra has an action of $C(1)$, so you can apply HH repeatedly and pass to a colimit to get something with an action of $C(\infty)$, which is an $E_\infty$ operad. If we are working over $\mathbb{Q}$ then $E_\infty$ algebras are essentially the same as commutative algebras. I think that this procedure converts free associative algebras to free commutative algebras.