The free group $F_2$ contains $F_k$

To elaborate on Martin's comment: let $x_i = ab^i a^{-1}$. Then what you have said is that $x_1 x_4 = x_5$ in what you are calling $F(S)$ (what you really mean is $\langle S\rangle$, the subgroup of $F_2$ generated by $S$). But $x_1 x_4$ is a reduced word in the $x_i$'s, so it shouldn't 'collapse to' a different word. Since it does, $\langle S\rangle$ is not free on $S$.

EDIT: Since what you were mainly asking about was whether your line of reasoning is correct, I'll elaborate a little more. When you define $F(S)$ to be the free group on $S$, then you are treating $S$ as an abstract set, rather than as a subset of $F_2$. This is perfectly valid in some sense, but it's not going to give you a subgroup of $F_2$, so it's not helpful.
What you want to do is to find, for each $k\in \mathbb{N}$, a subset $X_k$ of $F_2$ such that the subgroup generated by $X_k$ is free on $X_k$.

You're actually very close to the right idea, though. The standard way of doing this does involve taking conjugates, but remember that you don't want any 'collapsing' to occur except when you have something like $x_i x_i^{-1}$, so you'll need to do something a little differently.

Well, since there's another answer with an outline of a proof, and where to look etc., I'll add a partial spoiler for this way, but please don't look until you've tried.

Instead of setting $x_i = ab^i a^{-1}$, try $x_i = a^i b a^{-i}$. Now all that pesky 'collapsing' shouldn't be a problem. So if $S_k = \langle x_i \mid 1\leq i\leq k\rangle$, you should be able to show that $S_k\cong F_k$.

If $a,b$ generate $F_2$, then the family $\{c_k= a^k b a^{-k}, k \in \Bbb Z\}$ is free :

Look at a nontrivial word $c_{k_1}^{e_1}c_{k_2}^{e_2}\ldots c_{k_n}^{e_n}$ where $n>0$, $e_i \neq 0$ and $k_i \neq k_{i+1}$.
After translating, it is equal to $a^{k_1}b^{e_1}a^{k_2-k_1}b^{e_2}\ldots a^{k_n-k_{n-1}}b^{e_n}a^{-k_n}$. The only simplification that can possibly take place is if $k_1 = 0$ or $k_n = 0$, where the corresponding factor disappear, but you are still left with a nontrivial word in $a$ and $b$, so this cannot be the identity element.