Denote quadratic equations $E_1: z^2 + az + b = 0$, $E_2: z^2 + az + 2b = 0$. Let $z_1, z_2$ be solutions of $E_1$ and $z_3, z_4$ be solutions of $E_2$.
If $z_1, z_2, z_3, z_4$ are all real numbers, then their corresponding points on the complex plane are all on the real axis, which is not consistent with the question. If $z_1, z_2, z_3, z_4$ are imaginary numbers, then their corresponding points on the complex plane are all on line $\operatorname{Re} z = -\frac{a}{2}$, which does not fit the question. Therefore, there are two real numbers and two imaginary numbers in $z_1, z_2, z_3, z_4$.
This shows that discriminant $a^2-4b$ of equation $E_1$ and the discriminant $a^2-8b$ of $E_2$ have different signs.
At this point, there must be $b > 0$ (if $b \le 0$, then $a^2 - 4b \ge 0$ and $a^2 - 8b \ge 0$, a contradiction), so
$$a^2 - 4b \ge 0 > a^2 - 8b.$$
Hence, $z_{1,2} = \frac{-a \pm \sqrt{a^2 - 4b}}{2}$, $z_{3,4} = \frac{-a \pm \sqrt{8b - a^2}}{2}$.
It is evident that $\frac{z_1 + z_2}{2} = \frac{z_3 + z_4}{2} = -\frac{a}{2}$. Since the side length of the square is $1$, there is
$$|z_1 - z_2| = \sqrt{a^2 - 4b} = \sqrt{2},$$
$$|z_3 - z_4| = \sqrt{8b - a^2} = \sqrt{2},$$
namely, $a^2 - 4b = 8b - a^2 = 2$, and the solutions are $a^2 = 6, b = 1$.
Noticing that $z_1, z_2$ have the same sign and $|z_3| = |z_4|$, we know that
$$\begin{aligned} |z_1| + |z_2| + |z_3| + |z_4| &= |z_1 + z_2| + 2|z_3| \\ &= |-a| + \sqrt{a^2 + (8b - a^2)} \\ &= \sqrt{6} + 2\sqrt{2}. \end{aligned}$$