This will be CDF of discrete distribution with certain number of values $s_1,...,s_r$ where $r$ is number of possible distinct sums among all subsets $\{a_{i_1},...,a_{i_m}\}$. It is not possible to evaluate $r$ in general case because this number solely depends on particular values $a_1,...,a_n$ and can vary from $n+1$ to $2^n$. The minimal $r=n+1$ is represented by case of $a_1=a_2=...=a_n$ while maximal $r=2^n$ is always achieved when ascending reordered set $\{a_{(1)},...,a_{(n)}\}$ complies with the condition: $\forall i: \sum\limits_{1\le j<i}a_{(j)}\le a_{(i)}$. Example of such maximal case could be $a_i=2^{i-1}$. Note, that failing this condition does not necessarily mean $r<2^n$. For example, for $a_i=ln(prime(i))$ the condition above fails, however all possible sums $a_{i_1}+...+a_{i_m}$ are still distinct, causing $r=2^n$.
We can always consider set $\{s_1,...,s_r\}$ as ordered, then $s_1=0$ and $s_r=\sum\limits_i a_i$. Then $$P(s_1)=P(0)=\prod\limits_i (1-p_i)$$ as all $x_i=0$; $$P(s_r)=P\left(\sum\limits_i a_i\right)=\prod\limits_i p_i$$ as all $x_i=a_i$; and $$P(s_k)=\sum\limits_{i_1,...,i_m|a_{i_1}+...+a_{i_m}=s_k}\left(\prod\limits_{j\in\{i_1,...,i_m\} } p_j\times\prod\limits_{j\in[1;n]\setminus\{i_1,...,i_m\} } (1-p_j)\right)=\\P(0)\times\sum\limits_{i_1,...,i_m|a_{i_1}+...+a_{i_m}=s_k}\left(\prod\limits_{j\in\{i_1,...,i_m\} }\frac{p_j}{1-p_j}\right)$$ as we can eliminate the second product of complemented set $[1;n]\setminus\{i_j\}$ by using the fact that $P(0)$ contains all $(1-p_i)$ for the whole set $[1;n]$.
Best Answer
The answer $E_n$ depends on $n$, for $n=0$ it is clearly $E_0=0$. For $n\ge 1$, we have the recursion $$E_n=1 + \frac12 E_{n-1}$$ because with probability $\frac12$ we have $\min S=2$ and with probability $\frac12$ the result is the same as from taking a sample from $\{x_2,\ldots ,x_n\}=2\cdot\{x_1,\ldots,x_{n-1}\}$ and rejecting it with probabiliy $\frac12$. By induction, $E_n=2-2^{1-n}$.