Answer: NO. Let us argue by contradiction. Arrange the numbers in $A$ and $B$ in increasing order $1 \le a_1 < a_2 < \dots < a_{n+1} \le 2^n$ and $2^{n-1} < b_1 < b_2 \le 2^n$. Apparently, there exists an index $i \le n$ such that $a_i < a_{i+1} \le 2a_i$. We denote:
$$a_i b_1 - 1 = q_1^3,\quad a_{i+1} b_2 - 1 = q_2^3,\quad a_i b_2 - 1 = s_1^3,\quad a_{i+1} b_1 - 1 = s_2^3.$$
From $a_i < a_{i+1} \le 2a_i$ and $b_1 < b_2 < 2b_1$ it follows
$$q_1 < s_1,\quad s_2 < q_2 < 2q_1 \quad (1).$$

$$(q_1 q_2)^3 + q_1^3 + q_2^3 + 1 = (s_1 s_2)^3 + s_1^3 + s_2^3 + 1 \quad (2).$$

Using (1) it can be seen that if $q_1q_2 > s_1s_2$ or $q_1q_2 < s_1s_2$ the equality (2) cannot hold. Thus, $q_1q_2 = s_1s_2$. Let us consider the function $f(x) = x^3 + \frac{C}{x^3}$ in the interval $x \in [q_1, q_2]$, where $C = (q_1q_2)^3$. It decreases in $[q_1, \sqrt{q_1q_2}]$ and increases in $[\sqrt{q_1q_2}, q_2]$, hence it attains its maximum value at the ends of this interval, $f(q_1) = f(q_2) = q_1^3 + q_2^3$. Therefore, the equality (2), provided that (1) holds, is true only if $(q_1, q_2) = (s_1, s_2)$, which is the needed contradiction. $\Box$