One missing step in proving $\mathbb{Z}\times \mathbb{Z} \cong \langle a,b\,|\, [a,b]=1\rangle$

I think the issue is your approach. I'm sure there is a way to show that $\text{ker}(\pi) \subseteq N$, but as you've noticed, it is quite hard to argue this directly. The combinatorics of words can be extremely complicated and it can be useful to avoid talking about them for "simple" problems like this.

It is for exactly this reason that "abstract nonsense" exists. You are already halfway there by using the universal property of the free group, why not go all the way and use the universal property of quotients?

To show $F(S)/N \cong \mathbb{Z}^2$, let's show that $\mathbb{Z}^2$ satisfies the universal property of the quotient! That is, let's show that

• for any hom $f : F(S) \to G$
• if $N \subseteq \text{ker}(f)$
• then $f$ descends to a unique hom $\tilde{f} : \mathbb{Z}^2 \to G$

So let's fix a map $f : F(S) \to G$, which satisfies $N \subseteq \text{ker}(f)$. That is, $f([a,b]) = [f(a),f(b)] = 1$.

But then we can set $\tilde{f}((1,0)) = f(a)$ and $\tilde{f}((0,1)) = f(b)$. If you know that $\mathbb{Z}^2$ is the free abelian group on two generators, then we're done. If not, then you should check by hand that this is actually a homomorphism which makes the quotient diagram commute. I'm sure you'll find this much easier than trying to show that $\text{ker}(\pi) = N$. If you find yourself wanting another hint, feel free to ask in a comment.

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Edit:

To elaborate some, let's start with what the "universal property of the quotient" even means. The naming comes from a branch of math called Category Theory, and the idea is to characterize objects based on what morphisms they allow. You've already seen this in the Free Group, which has a universal property expressing what morphisms from the free group exist.

We can describe lots of different algebraic constructions in this way, including the quotient! The definition is as follows:

If $N \trianglelefteq G$, then a morphism $\pi : G \to K$ satisfies the universal property of the quotient (with respect to $G$ and $N$) if and only if: For every $f : G \to H$ with $N \subseteq \text{ker}(f)$, there is a (unique!) homomorphism $\tilde{f} : K \to H$ such that $f = \tilde{f} \circ \pi$.

Now I will leave it to you to show the following few facts, which will culminate in the theorem you want to prove:

1. Show $\pi : G \to G/N$ the natural quotient map satisfies the universal property of the quotient.

   • This says that the name is well chosen. Indeed, this is where the name comes from!

2. If two maps $\pi_1 : G \to K_1$ and $\pi_2 : G \to K_2$ both satisfy the universal property, then $\varphi : K_1 \cong K_2$, and moreover $\pi_2 = \pi_1 \circ \varphi$. So not only are $K_1$ and $K_2$ "the same", but so are the quotient maps $\pi_1$ and $\pi_2$.

   • If you haven't seen proofs like this before, this might take some ingenuity. As a hint, notice $\pi_2$ is a map out of $G$ with $N \subseteq \text{ker}(\pi_2)$. So since $K_1$ has the universal property, we must have a map $K_1 \to K_2$. Swapping the roles of $K_1$ and $K_2$, we also have a map $K_2 \to K_1$... Can you prove these maps compose to the identity?

3. Now can we show that $\mathbb{Z}^2$ satisfies the universal property with respect to $G = F(a,b)$ and $N = \langle [a,b] \rangle$?

   • This is what I outlined in the original part of the answer.

4. Finally, since $F(a,b)/N$ and $\mathbb{Z}^2$ both satisfy the universal property (by (1.) and (3.) respectively), conclude they must be isomorphic (by (2.)), which is what we're trying to show.

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I hope this helps ^_^