Since the polynomial $f(x) = x^3 + a x^2 + 2b x - 1$ has three different roots, the equation
$$x^3 + a x^2 + 2b x - 1 = 0 \quad (1)$$
has three different solutions, denote them by $x_1$, $x_2$ and $x_3$. Vieta's formulas implies the equalities $x_1 + x_2 + x_3 = -a$, $x_1 x_2 + x_1 x_3 + x_2 x_3 = 2b$ and $x_1 x_2 x_3 = 1$. Hence
$$x_1^2 + x_2^2 + x_3^2 = (x_1 + x_2 + x_3)^2 - 2(x_1 x_2 + x_1 x_3 + x_2 x_3) = a^2 - 4b.$$
From the AM-GM inequality:
$$a^2 - 4b = x_1^2 + x_2^2 + x_3^2 > 3 \sqrt{x_1^2 x_2^2 x_3^2} = 3 \quad (2)$$
(the inequality is strict, since the solutions of (1) are distinct).

The quadratic polynomial $g(x)$ doesn't have roots, therefore its discriminant is negative, i.e. $4b^2 - 8a < 0$, whence $2a - b^2 > 0$. Summing this inequality with (2) we obtain $(2a - b^2) + (a^2 - 4b) > 3$, which is equivalent to $(a + 1)^2 - (b + 2)^2 + 3 > 3$. Therefore,
$$(a + 1 + b + 2)(a + 1 - b - 2) > 0.$$
Since $a$ and $b$ are positive, $a + b + 3 > 0$, so the last inequality implies $a - b - 1 > 0$.