If two spaces are homeomorphic, then their fundamental groups are isomorphic. Conversely, if two spaces have non-isomorphic fundamental groups, then they cannot be homeomorphic to each other. So to prove that two spaces are not homeomorphic, it's enough to show that their fundamental groups are not isomorphic to each other.
Assume $G$ is a non-abelian group and $H$ is an abelian group. Assume $f:G \to H$ were an isomorphism. Choose $x, y \in G$ such that $xy \neq yx$. Then because $f$ is a homomorphism, $f(xy)=f(x)f(y)$. Since $H$ is abelian, $f(x)f(y)=f(y)f(x)=f(yx)$. And because $f$ is an isomorphism, $f(xy)=f(yx) \Rightarrow xy=yx$, contradicting our assumption that $xy \neq yx$. Thus, no such isomorphism $f$ can exist.
You are correct, the existence holds, but the uniqueness statement is false in several regards.
First of all, it is true that $\Bbb Z^2$ as a $\Bbb Z$-module can be expressed as, for instance $(1) \oplus (1)$ as well as $(1) \oplus (2)$. More generally, you could observe that the statement contradicts its analogue for PID, which says that finitely generated projective modules over a PID $A$ are free, i.e. isomorphic to direct sums of the unit ideal: take any non-unit ideal of $A$ as a counterexample for the uniqueness of Lang's statement.
Seeing this argument, you might think that perhaps there is a way to salvage uniqueness, by imposing that the ideals $\mathfrak{a}$ are taken to be as big as possible: any principal ideal of $A$ is isomorphic (as an $A$-module) to the unit ideal and is contained inside of it ; if we replace principal ideals by the unit ideals in our argument for PID's, we do find uniqueness.
Unfortunately, the answer is still no, and here is a counterexample. Consider $A = \Bbb Z[\sqrt{-5}]$, which is known to be a non-principal Dedekind domain. Let $M=\mathfrak{p} := (2,1+\sqrt{-5})$, let $N = \mathfrak{q} := (3,1-\sqrt{-5})$. These are distinct maximal ideals of $A$, so neither is contained in the other. However, the map
$$m \mapsto \frac{m \times 3}{1+\sqrt{-5}}$$
defines an isomorphism of $A$-modules from $\mathfrak{p}$ to $\mathfrak{q}$ (it suffices to check the images of generators).
You can stop here if you just wanted a definitive counterexample. I will now carry on to show where Lang's proof fails.
Lang shows existence as follows. Pick any nonzero $\lambda \in \operatorname{Hom}_A(M,A)$, which exists (pick a nonzero linear map $M \otimes_A K \rightarrow K$, where $K= \operatorname{Frac}(A)$, and multiply by the common denominator of the images of the generators of $M$). Denote $\mathfrak{a}_1 := \lambda(M)$, it is a finitely generated torsion-free $A$-module, hence projective (Lang Proposition 26), so the following short exact sequence splits:
$$0 \rightarrow M_1 := \ker \lambda \rightarrow M \rightarrow \mathfrak{a}_1 \rightarrow 0.$$
Hence we can write $M \simeq \mathfrak{a}_1 \oplus M_1$. By tensoring with $K$, one sees that the rank of $M_1$ is less than the rank of $M$, so one would be tempted to carry on by induction. But doing so, we would have no way of imposing the division relation between the ideals. Instead, if we first pick $\lambda$ as above so that the image is as large as possible (I'm avoiding the term "maximal" because the image need not be a maximal ideal), then we are guaranteed that the remaining ideals are contained in $\mathfrak{a}_1$. To see this, assume that $M_1 \simeq \mathfrak{a}_2 \oplus M_2$ with $\mathfrak{a}_2$ not contained in $\mathfrak{a}_1$, then $M \simeq \mathfrak{a}_1 \oplus \mathfrak{a}_2 \oplus M_2$, from which we find a map $\lambda':M \rightarrow A$ whose image contains both $\mathfrak{a}_1$ and $\mathfrak{a}_2$, contradicting the maximality of $\lambda$. This concludes the existence.
For uniqueness, Lang simply says
The uniqueness follows by localizing at primes $\mathfrak{p}$, and invoking the corresponding uniqueness over principal rings, which is part of standard algebra.
However, this does not have the desired effect. If we write
$$M = \bigoplus_{s=1}^t \mathfrak{p_1}^{e_{1,s}}...\mathfrak{p_m}^{e_{m,s}}$$
where each $(e_{j,s})_s$ is a nondecreasing sequence of nonnegative integers (this is a reformulation of the existence), localizing at (the complement of) $\mathfrak{p}$ gives:
$$M_{(\mathfrak{p})} = \bigoplus_{s=1}^t A_{(\mathfrak{p})}\mathfrak{p_1}^{e_{1,s}}...\mathfrak{p_m}^{e_{m,s}}.$$
Now each term of the sum is an ideal of $A_{(\mathfrak{p})}$ (the unit ideal if $\mathfrak{p}$ is not one of the factors), hence principal, hence isomorphic to a free module of rank 1, and we have effectively lost all information about which ideals were there, and to which power. Only the rank of $M$ remains.
In the proof of existence, it might happen that there exists two choices for a maximal $\lambda$ which map to distinct, incomparable ideals $\alpha$, in which case uniqueness would fail. This is how I found the counterexample above.
Best Answer
The missing relations are $\alpha^2 = 2 - 2(4/\alpha) - \alpha$ and $(4/\alpha)^2 = 4/\alpha - 2\alpha - 2$.
These translate into $x^2 = 2 - 2y - x$ and $y^2 = y - 2x - 2$.
Modulo $2$ they become $x^2 = x$ and $y^2 = y$ (and these make the cubic relations superfluous).
So you have $$\begin{align*}\mathbb{F}_2[x,y]/(x^2 - x, xy, y^2 - y) &= \mathbb{F}_2[x,y]/(x,xy,y^2-y) \cap (x-1,xy,y^2-y) \\&\cong \mathbb{F}_2[x,y]/(x,xy,y^2-y)\oplus \mathbb{F}_2[x,y]/(x-1,xy,y^2-y)\\ & \cong \mathbb{F}_2[y]/(y^2-y)\oplus \mathbb{F}_2[y]/(y,y^2 - y)\\ &\cong \mathbb{F}_2 \oplus\mathbb{F}_2 \oplus\mathbb{F}_2.\end{align*}$$