I'm having trouble finding the error in my proof:
Theorem: If H is a normal subgroup of G, then the quotient group G/H is abelian.
Proof: $\forall x, y \in G$, $xy(yx)^{-1} = xyy^{-1}x^{-1} = e \in H$. By the lemma, $H(xy) = H(yx)$, thus $HxHy= HyHx$. This shows that the commutative property holds for all the elements in the quotient group G/H, therefore G/H is abelian.
Lemma: Let H be a subgroup of G. Then $Ha = Hb \iff ab^{-1} \in H$. Proof: We start with the $\Rightarrow$ direction. $Ha = Hb$ implies $a \in Hb$, thus $a=hb$ for some $h \in H$. Therefore $ab^{-1} = h \in H$. For the reverse direction, if $ab^{-1} = h \in H$ then $a=hb$ for some $h \in H$. Thus $a \in Hb$, and since we also have $a \in Ha$, $Ha = Hb$ (because cosets partition a group).
Look at the first line of your proof: $xy(yx)^{-1}$ is not $xyy^{-1}x^{-1}$ (unless $G$ is itself abelian). Instead $xy(yx)^{-1}=xyx^{-1}y^{-1}$, which is not guaranteed to simplify to $e$.
Bringing in a bit of jargon, the point is that the "inversion" map $$G\rightarrow G: g\mapsto g^{-1}$$ is only guaranteed to be an "antihomomorphism" instead of a homomorphism: we have $(ab)^{-1}=b^{-1}a^{-1}$, and you can see that by multiplying $$(b^{-1}a^{-1})(ab)=b^{-1}(a^{-1}a)b=b^{-1}b=e.$$ In fact, the statement "$(ab)^{-1}=a^{-1}b^{-1}$" (or "inversion is a homomorphism") is equivalent to the abelian-ness of the group in question. Essentially, this is what your argument actually shows: if your first sentence is justified, then indeed the group in question is abelian.