Let’s assume the contrary. We can add some singletons to the group of sets $A$ to make $f(\{x\}) = g(\{x\})$ for all $1\leq x\leq n$. Each singleton is added at least once. Assume the total number of sets in group $A$ is $p$. Let $v_1, \ldots, v_p \in \mathbb{R}^n$ be the incident vectors of sets $A_1, \ldots, A_p$ such that the $j^{th}$ entry of $v_i$ is $1$ if $j \in A_i$, and $0$ otherwise. Similarly define $u_1, \ldots, u_t$ be the incident vectors of sets $B_1, \ldots, B_t$. Define matrices $A=[v_1v_2\ldots v_p]$ and $B=[u_1u_2\ldots u_t]$. Since $f(\{x\}) = g(\{x\})$ and $f(\{x, y\}) = g(\{x, y\})$ for all $x, y$, we have $AA^T=BB^T$. However, $rank(AA^T)=n$ since $A$ has all singletons, but $rank(BB^T)\leq min(rank(B^T), rank(B)) \leq t <n$, contradiction! Hence we’re done.