If $G$ is a finite group and $|G| < |A| + |B|$, then $G=AB$.
Let $g \in G$. Note that $|gB^{-1}| = |B|$ so $|A| + |gB^{-1}| > |G|$; in particular, $A\cap gB^{-1} \neq \emptyset$. Let $a \in A\cap gB^{-1}$, then $a = gb^{-1}$ for some $b \in B$. Therefore $g = ab \in AB$ so $G = AB$.
Let $A = \{ a_1, \ldots, a_n \}$, and let $g \in G$ be arbitrary. Let $x_k = a_k^{-1} g$. Then $x_1, \ldots, x_n$ are distinct elements of $G$, and since the complement of $B$ in $G$ has less than $n$ elements, there exists $k$ such that $x_k \in B$. Then $g = a_k x_k \in AB$, so $G \subseteq AB$, proving the claim.