I agree that it is not an variety because we cannot write it as the variety defined by the polynomial $y-bx$, since here the $b$ is not fixed. I think there is a key point is that we get two parameters instead of only a single parameter, so we cannot fix a set of polynomial such that the set $\{(a,ab) | a,b ∈ \Bbb C^2\}$ is exactly its solution set. But I wonder how I can give a formal proof that it cannot be zero locus for any set of polynomials(perhaps by contradiction, I think). May I please ask for a proof? Any references are also appreciated.
Thanks in advance.
EDIT: Now I am also wondering if I can prove it without using the fact that this set is not closed. If such a method exists, could someone let me know? Thanks.