With the added information, which is key (otherwise, the two expressions are just not equal in general, as noted by Ross):
If $b=1-a$ and $b\gt a$, then $1-a\gt a$ so $1\gt 2a$. Then $1 -4ab = 1-4a(1-a) = 1-4a+4a^2 = (1-2a)^2,$ and since $1-2a\gt 0$, we have $\sqrt{1-4ab} = \sqrt{(1-2a)^2} = |1-2a| = 1-2a.$ So $\begin{align*} \frac{1+\sqrt{1-4ab}}{2a} &= \frac{1+(1-2a)}{2a} = \frac{2(1-a)}{2a} = \frac{b}{a},\\ \frac{1-\sqrt{1-4ab}}{2a} &= \frac{1-(1-2a)}{2a} = \frac{2a}{2a} = 1. \end{align*}$ That means that $\begin{align*} \left({1+\sqrt{1-4ab}\over2a}\right)^n-\left({1-\sqrt{1-4ab}\over2a}\right)^n\over \left({1+\sqrt{1-4ab}\over2a}\right)^m-\left({1-\sqrt{1-4ab}\over2a}\right)^m & = \frac{\left(\frac{b}{a}\right)^n - 1^n}{\left(\frac{b}{a}\right)^m - 1^m}\\ &=\frac{\left(\frac{b}{a}\right)^n - 1}{\left(\frac{b}{a}\right)^m - 1}. \end{align*}$