Assume $a_1 \le a_2 \le \dots \le a_n$. If we choose the subset $B' = \{1, 2, \dots, 2^{n-2}\}$, then among the sums $s_1, s_2, \dots, s_{2^{n-2}}$, the values $a_1, a_2, \dots, a_t$ will appear for some $t \ge 1$. Clearly, $s_1 = s_{\{1\}} = a_1$, and if $a_i$ appears as a sum, then $a_{i-1}$ must have appeared before it.
Indeed, consider:

a) if $a_{i-1} = a_i$, then the appearance is obvious.

b) if $a_{i-1} < a_i$, then since $s_{\{i-1\}} = a_{i-1} < a_i$ and $a_i$ appears as some $s_j$, the index corresponding to $a_{i-1}$ is less than $j$, hence $a_{i-1}$ appeared earlier.

If $t \le n-2$, then at most $2^{n-2}-1$ sums can appear, corresponding to non-empty subsets of $\{1, 2, \dots, n-2\}$. Hence, we deduce that $t \ge n-1$.
Therefore, the first two sums correspond to the numbers $a_1$ and $a_2$. Then, inductively, after determining the numbers $a_1, a_2, \dots, a_k$, we compute all the subset sums of $\{a_1, a_2, \dots, a_k\}$, remove them from the list of known sums, and the smallest remaining sum must be $a_{k+1}$.
So, we can determine the numbers $a_1, a_2, \dots, a_{n-1}$.
If we also include the index $2^n - 1$, we obtain $s_{2^n-1}$, which is the total sum of all numbers. Then we can compute $a_n$, and thus all numbers.