The statement is true, assuming that by the abelianization $G^{ab}$ of a profinite group $G$ we mean its abelianization as a profinite group: $G^{ab} = G/[G,G]$, where $[G,G]$ is the closed subgroup generated by all commutators in $G$.
Let's write $G = G(\overline{K}\mid K)$. We want to show that $G^{ab}$ is the Galois group of $K^{ab}$ over $K$. Of course, since $G^{ab}$ is a quotient of $G$, the usual Galois theory for fields tells us that $G^{ab}$ is the Galois group of a well-defined field $L$ with $K \subseteq L \subseteq \overline K$, namely the fixed field in $\overline K$ of the commutator subgroup $[G, G]$ of $G$. So we want to show that $K^{ab} = L$.
For this, first notice that $L$ is an abelian extension of $K$, since by construction its Galois group $G^{ab}$ is abelian. Since $K^{ab}$ is the maximal abelian extension of $K$, this means that $L \subseteq K^{ab}$. Therefore, by the Galois correspondence, we get the reverse inclusion on the side of groups, namely $G(\overline K\mid K^{ab}) \subseteq G(\overline K \mid L) = [G, G]$.
On the other hand, since $K^{ab}$ is an abelian extension, any commutator in $[G, G]$ must fix $K^{ab}$ element-wise, so we also get $[G, G] \subseteq G(\overline K \mid K^{ab})$. Combined with the previous inclusions this shows that $G(\overline K\mid K^{ab}) = G(\overline K \mid L)$, so $K^{ab}$ and $L$ correspond to the same subgroup of $G$, and again using the Galois correspondence then shows that $K^{ab} = L$.
