Small world theorem for set constraints
The conjecture is true, and the proof strategy listed there can be made to work.

Let $s = (s_1,\dots,s_n)$ be a satisfying assignment for the variables that makes all of the constraints hold.  We'll write $E(s)$ for the result of evaluating the expression $E$ with the variables $S_i$ set to $s_i$.  In the following, define $U$ by the procedure listed in the question:  let $E_1 \ne \emptyset$, ..., $E_k \ne \emptyset$ be the constraints of second type; pick any element of $E_1(s) \cup \cdots \cup E_k(s)$, call it $x$; add $x$ to $U$; delete all of the constraints $E_i \ne \emptyset$ such that $x \in E_i$; and repeat.  Define $s'_i = s_i \cap U$.

Now I claim that $s'_1,\dots,s'_n$ is also a satisfying assignment that makes all of the constraints hold.

First, let's consider the constraints of the form $E = \emptyset$.  I claim that if $E(s) = \emptyset$, then $E(s') = \emptyset$.  In particular, I claim that $E(s) \cap U = E(s') \cap U$ for all set expressions $E$.  The proof is by recursion: by definition of $s'$, we have $S_i(s) \cap U = S_i(s') \cap U$; also if $t \cap U = t' \cap U$, then $\overline{t} \cap U = \overline{t'} \cap U$; and if $t_1 \cap U = t'_1 \cap U$ and $t_2 \cap U = t'_2 \cap U$, then $(t_1 \cap t_2) \cap U = (t'_1 \cap t'_2) \cap U$.  So this means that $s'$ satisfies all of the constraints of the first type (all of the constraints of the form $E = \emptyset$), since $s$ does.

Next, consider the constraints of the form $E \ne \emptyset$.  We know $E(s) \ne \emptyset$.  Therefore, there was a step in the iterative definition of $U$ when we chose some element $x \in E(s)$ and added it to $U$.  We proved earlier that $E(s) \cap U = E(s') \cap U$.  Since $x \in E(s)$ and $x \in U$, it follows that $x \in E(s') \cap U$, and thus $E(s') \ne \emptyset$.  This means that $s'$ satisfies all of the constraints of the second type, too.

We've shown that if there is any satisfying assignment for all constraints, then there exists a satisfying assignment where all sets are subsets of $U$.  This completes the proof of the conjecture.