The answer is no. There are uniquely determined linear polynomials $a_m(x):=x-c_m$ such that $g(x)$ is identically zero on $\mathbb{Q}$. Indeed, the condition says that, for any $n\geq 1$,
$$ \sum_{0\leq m\leq n-1}(b_n-c_m)\cdot(b_n-b_1)\dots(b_n-b_m) = 0. $$
Suppose that $c_m$ has been chosen already for every $0\leq m<n-1$. Then the above equation determines $c_{n-1}$ uniquely, and we are done.