Let $A\subseteq B$ be two integral domains such that the multiplication function $m: A \times (B \setminus A) \to B$, $m(x,y)=xy$, is surjective.
Let $S \subset A[x]$ be the set of all monic polynomials. For each $b \in B$, let ${v}_b: A[x] \to B$ be the evaluation homomorphism, given by $v_b(f) = f(b)$.
Prove that there exists $b_0 \in B$ such that $({v}_{b_0}(S))^{-1}B \neq \{0\}$.
We look for some $b_0,b \in B, f \in S$ such that $\frac{b}{f(b_0)} \neq \frac{0}{1}$. Thus we look for some elements $b_0, b \in B$ such that for all $g \in S$, $b\cdot g(b_0)\neq 0 $ in $B$. Since $B$ is an integral domain, this is equivalent to requiring that both $b, g(b_0)$ are nonzero. Hence the choice of $b$ doesn't seem to matter, and we only want to find a suitable $b_0 \in B$ that is not the root of any monic polynomial in $A[x]$.
Clearly we must have $b_0 \in B \setminus A$, since otherwise we can take $g = x-b_0 \in S$ and get $b\cdot g(b_0) = 0 $ no matter how we choose $b$.
We didn't use our given information that $m$ is surjective and $A$ is an integral domain. The surjectivity implies that there are $x \in A, y \in B \setminus A$, with $xy=1$. Thus $x,y$ are units in $B$. Which seems to be a dead-end.
Let me start form here: The surjectivity implies that there are $x \in A, y \in B \setminus A$, with $xy=1$.
Suppose that $A\subset B$ is an integral extension. Then $y$ is integral over $A$ and write $$y^n+a_1y^{n-1}+\cdots+a_n=0,$$ where $a_i\in A$. Multiply this by $x^n$ and find $1+a_1x+\cdots+a_nx^n=0$, so $x$ is invertible in $A$ hence $y\in A$, a contradiction.