Combining the fractions on each side, we get \[\frac{ax - a^2 + bx - b^2}{ab} = \frac{ax - a^2 + bx - b^2}{(x - a)(x - b)}.\]Note that the numerators are equal. The solution to $ ax - a^2 + bx - b^2 = 0 $ is \[x = \frac{a^2 + b^2}{a + b}.\]Otherwise, \[\frac{1}{ab} = \frac{1}{(x - a)(x - b)},\]so $ (x - a)(x - b) = ab $. Then $ x^2 - (a + b) x + ab = ab, $ so $ x^2 - (a + b) x = 0 $. Hence, $ x = 0 $ or $ x = a + b $. Thus, there are $ \boxed{3} $ solutions, namely $ x = 0, $ $ x = a + b, $ and $ x = \frac{a^2 + b^2}{a + b} $. (If $ \frac{a^2 + b^2}{a + b} = a + b, $ then $ a^2 + b^2 = a^2 + 2ab + b^2, $ so $ 2ab = 0 $. This is impossible, since $ a $ and $ b $ are non-zero, so all three solutions are distinct.)