Combinations of set unions
I understood the problem as that we can select any $S_i$ satisfying these conditions. Then the answer is $\{0, 1, 2, 3, 4, 5\}$. The construction is unique in a certain sense.
Wlog we can assume that $S=A$ (otherwise, just remove redundant elements from $S$). I'll use $A, B, C, D, E$ instead of $S_i$. One way to arrive at a solution is to draw a Venn diagram for $A, B, C$ and to think about it:

What can we say about it:

$A \setminus (B \cup C) \ne \emptyset$. Otherwise $B \cup C = A \cup B \cup C = S$. Similarly for $B$ and $C$.

For $D$ we must have $A \setminus (B \cup C) \subseteq D$. Otherwise, $B \cup C \cup D \ne B \cup C \cup A = S$. Similarly for $B$ and $C$.

For $D$ we must have $A \cap C \setminus B \not \subseteq D$. Otherwise (just look at the diagram),
$$B \cup D = B \cup (A \setminus (B \cup C)) \cup (C \setminus (A \cup B)) \cup (A \cap C \setminus B) = S$$

What we said for $D$ also holds for $E$. I.e. it must contain the symmetric difference but must not contain $A \cap C \setminus B$ and similar.

We must have $D \cup E \cup X = S$ for $X = A,B,C$. Therefore, $D \cup E \cup (A \cap B \cap C) = S$. In particular, it means that $A \cap B \cap C \not \subseteq D \cup E$, since otherwise $D \cup E = S$.
Also, $S \setminus (A \cap B \cap C) \subseteq D \cup E$. Therefore, forgetting about symmetric difference:
$$((A \cap C) \cup (A \cap B) \cup (B \cup C)) \setminus (A \cap B \cap C) \subseteq D \cup E$$
Therefore, while $A \cap C \setminus B$ doesn't belong to either $D$ or $E$, it belongs to their union.


Note that conditions above are necessary and sufficient. Putting this together:

$A \setminus (B \cup C)$ must have at least one element, which also belongs to both $D$ and $E$. The same for $B, C$.
$A \cap C \setminus B$ must have at least two elements: one belongs to $D$, another belongs to $E$. The same for other pairwise intersections.
$A \cap B \cap C$ has at least one element, which doesn't belong to either $D$ and $E$.

Note that we've used all $10$ elements from $S$. All sets $A,..,E$ have $6$ elements, so we just select one of them to be $S_1$, and arrange elements in the sets so that $S_1 = \{0,..,5\}$