Note that $x(x+1)^3 - (2x+a)(x+a-1) = x^4 + 3x^3 + x^2 + x(3-3a) + a - a^2 = (x^2 + 3x + a)(x^2 + 1 - a) = 0$.

In order to get 4 roots the discriminants of both quadratic polynomials must be positive: $9 - 4a > 0$ and $4a - 4 > 0$.

The roots $x^2 + 3x + a = 0$ and $x^2 + 1 - a = 0$ should be distinct: let $t$ be a common root, then $(t^2 + 3t + a) - (t^2 + 1 - a) = 3t + 2a - 1 = 0$ and $t = \frac{1-2a}{3}$.

Putting the last expression into $t^2 + 1 - a = 0$ we get $4a^2 - 13a + 10 = (4a-5)(a-2) = 0$.

Direct check shows that at $a = 2$ and $a = \frac{5}{4}$ two roots coincide.

Answer: $a \in (1, \frac{5}{4}) \cup (\frac{5}{4}, 2) \cup (2, \frac{9}{4})$.