We want to show that $I_AA'$ is the radical axis of $(I_ABC')$ and $(I_ACB')$, so it suffices to show that $A'$ has equal power with respect to these two circles. Let $X = BC \cap (I_ABC')$; then $\angle{I_AXB} = \angle{I_AC'B}$, meaning that a spiral similarity at $I_A$ takes $C'$ to $X$ and the tangency point of the $A$-excircle on $AB$ to its tangency point on $BC$, meaning that the distance from $X$ to the foot from $I_A$ to $BC$ is equal to the distance between the feet from $I_A, I$ onto $AB$, which is $BC$. Thus, $Pow_{(I_ABC')}(A') = 2 \cdot A'B \cdot A'C = Pow_{(I_ACB')}(A')$. $\square$

Let $N$ be other intersection of $I_AA'$ with incircle. Note from $\angle C'NB' = \angle C'B'A' = 90 - \frac{\angle B}{2}$ and $\angle I_aBC' = 90 + \frac{\angle B}{2}$ we have $I_a,B,C',N$ cyclic. Similary $I_a,C,B',N$ cyclic. $NI_a$ is radical axis of $(I_aBC')$ and $(I_aCB')$. Hence we have our statement true.