Your notation $S^*$ is a bit misleading. So let me first restate your question (as I understand it). Let $A = \{a,b\}$, $L = A^* \setminus \{a, b, ab\}$ and let
$$
S = \{(u_1, v_1), (u_2, v_2), \ldots, (u_n, v_n) \}
$$
be a finite set of pair of words of $A^*$. You are asking for some sufficient condition on $S$ to guarantee that
$$
L \subseteq \{u_1, v_1\}^* \cup \{u_2, v_2\}^* \cup \dotsm \cup \{u_n, v_n\}^*
$$
Theorem. This happens if and only if one of the sets $\{u_i, v_i\}$ is equal to $A$.
Proof. The condition is clearly sufficient. To prove it is also necessary, first recall that a subset $P$ of $A^*$ is dense if, for every word $w \in A^*$, $A^*wA^* \cap P \not= \emptyset$. It is nondense otherwise. For instance, $L$ is dense.
Step 1. Let $u, v \in A^*$. Then $\{u,v\}^*$ is dense if and only if $\{u,v\} = A$.
Let $n = 1 + 2\max\{|u|, |v|\}$. If $\{u,v\}^*$ is dense, there exist two words $x$ and $y$ such that $xa^ny \in X^*$. It follows that $u$ or $v$ is of the form $a^i$ for some $i > 0$. A similar argument shows that $u$ or $v$ is of the form $b^j$ for some $j > 0$. Thus $\{u, v\} = \{a^i, b^j\}$. I let you now verify that $\{u, v\}^*$ is dense if and only if $i = j = 1$.
Step 2. If $\{u_1, v_1\}^*, \{u_2, v_2\}^*, \dotsm \{u_n, v_n\}^*$ are nondense, then so is $\{u_1, v_1\}^* \cup \{u_2, v_2\}^* \cup \dotsm \cup \{u_n, v_n\}^*$.
Indeed, if $\{u_1, v_1\}^* \cap A^*w_1A^* = \emptyset$, $\{u_2, v_2\}^* \cap A^*w_2A^* = \emptyset$, $\dots$, $\{u_n, v_n\}^* \cap A^*w_nA^* = \emptyset$, then
$$
\bigl(\{u_1, v_1\}^* \cup \{u_2, v_2\}^* \cup \dotsm \cup \{u_n, v_n\}^*\bigr) \cap A^*w_1w_2 \dotsm w_nA^* = \emptyset
$$
Final step. If none of the sets $\{u_i, v_i\}$ is equal to $A$, then the sets $\{u_i, v_i\}^*$ are nondense by Step 1, and so is their union by Step 2. Thus this union cannot contain the dense set $L$.
