Utilize your discriminant!

We first rewrite the equation as

$\frac{x(x-b) + x(x-a)}{(x-a)(x-b)}= 1$

Which we can simplify to give

$x^{2} - bx + x^{2} - ax = x^{2} - ax - bx + ab$

$x^{2} - ab = 0$

If this equation has 2 distinct real solutions then $B^{2} - 4AC > 0$

Therefore

$0^{2} - 4 \times 1 \times -ab > 0$

$4ab > 0$

Thus $4ab$ must be strictly positive . This can only occur when $a$ and $b$ are both positive or when $a$ and $b$ are both negative.

$\begin{aligned} \dfrac{x}{x-a}+\dfrac{x}{x-b}&= 1\\ {(x)}{(x-b)}+{(x)}{(x-b)}&={(x-a)}{(x-b)} \\ x^2-ax+x^2-bx&=x^2-ax-bx+ab \\ \implies x^2&=ab \end{aligned}$

Since $x$ is a real number, $ab>0$ , which means both $a$ and $b$ must have the same signs.