For some reason the OP won't post, or can't post, an answer, so summarizing the comments:
$(1)\,\,\forall\,g,x,y\in G\,$ , and putting $\,a^b:=b^{-1}ab\,\,,\,a,b\in G\,$:
$$[x,y]^g:=g^{-1}[x,y]g:=g^{-1}x^{-1}y^{-1}xy g=\left(x^{-1}\right)^g\left(y^{-1}\right)^gx^gy^g=[x^g,y^g]\in G'\Longrightarrow G'\triangleleft G$$and thus the quotient $\,G/G'\,$ is a group.
$(2)\,\,$ Let now $\,N\,$ be any normal subgroup of $\,G\,$ s.t. $\,G/N\,$ is abelian, then:
$$\forall\,x,y\in G,\,\,xNyN=yNxN\Longleftrightarrow xyN=yxN \Longleftrightarrow (yx)^{-1}xy\in N \Longleftrightarrow [x,y] \in N$$
and since $\,G':=\langle\,[x,y]\;:\;x,y\in G\,\rangle\,$ , then $\,G'\leq N\,\Longrightarrow \,G'$ is the minimal (normal) subgroup of
$\,G\,$ s.t. its quotient is abelian -- "minimal" wrt set inclusion --.
Exercise: Explain the parentheses around "normal" above, i.e. show that any subgroup of G containing the commutator subgroup is normal.
Denote the commutator of $a$ and $b$ by $a^{-1}b^{-1}ab = [a,b]$.
If $u$ is an element from the commutator subgroup, then $g^{-1}ug = u(u^{-1}g^{-1}ug) = u[u, g]$ .
Another approach: the commutator subgroup is defined to be the subgroup generated by the commutators, so every element of the commutator subgroup is of the form $$[a_1, b_1][a_2,b_2]\ldots[a_n, b_n].$$ It is enough to show that $g^{-1}[a,b]g$ is always in the commutator subgroup, because then
$$g^{-1}[a_1, b_1][a_2,b_2]\ldots[a_n, b_n]g = (g^{-1}[a_1, b_1]g)(g^{-1}[a_2,b_2]g)(g^{-1}\ldots g)(g^{-1}[a_n, b_n]g)$$
is a product of elements from the commutator subgroup. When $\phi$ is any homomorphism, we have $\phi([a,b]) = [\phi(a), \phi(b)]$. Since for any $g \in G$ the map $\phi$ defined by $\phi(x) = g^{-1}xg$ is a homomorphism, the result follows.
Best Answer
$G/H=\{gH: g\in G\}$ by definition. this is only a group under $(gH)(g'H) = (gg')H$ if $Hg' = g'H$. But this is just another way of stating the definition of $H$ being normal. In your proof you just neglected to note that $xyx^{-1}y^{-1}H$ is only relevant because it is equal to $(xH)(yH)(x^{-1}H)(y^{-1}H)$ because $H$ is normal.
I would call this "incomplete" rather than "wrong" if anything, as the problem is a few steps beyond reproving the basic fact that $G/H$ is only a group when $H$ is normal. I think you just forgot that that's what makes $G/H$'s group operation well-defined.