Let $a_i>b_i$ for all $i$ without loss of generality. Note that \[ \{a_1+b_1\} + \{a_2,b_2\} = \{a_1+a_2,a_1+b_2,b_1+a_2,b_1+b_2\}, \]and \[ a_1+a_2 > \max\{a_1+b_2,b_1+a_2\} > \min\{a_1+b_2,b_1+a_2\} > b_1+b_2. \]We this immediately deduce that \[ a_1 + a_2 = a_3 + a_4 \quad\text{and}\quad b_1+b_2 = b_3 + b_4. \]There are now two cases to consider. Case 1. $a_1+b_2 = a_3+b_4$ and $a_2+b_1 = a_4+b_3$. We then have $a_1-b_1 = a_3-b_3$ and $a_2-b_2 = a_4-b_4$, so \[ \prod_{1\le i\le 4}|a_i-b_i| = |a_1-b_1|^2 \cdot |a_2-b_2|^2, \]which is perfect square. Case 2. $a_1+b_2 = a_4+b_3$ and $a_2+b_1 = a_3+b_4$. Handled analogously.