Essentially what the question title says, $ab$ is just usual multiplication.
Is $P(ab \mid a) = P(b)$? This intuitively seems true to me since if we fix $a$ then the product $ab$ strictly depends on what $b$ turns out to be, i. e., $P(b)$?
Edit:
Consider two 10 faced die where faces are $F = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$. Let's say a pair of die are rolled one after another and the first one comes out $a \in F$. What is the probability that once the second is also rolled, the product of the two faces is $ab$ given the first roll was $a$. This is to say $P(ab \mid a)$. If I am right then, $$ P(ab \mid a) = P(a \mid a) \cdot P(b \mid a) = P(b \mid a) = P(b) $$ The last equality is due to statistical independence.
Trying to generalize your example, suppose you have two independent discrete random variables $X$ and $Y$. You want to know if $\mathbb P(XY = xy \mid X = x) = \mathbb P(Y = y)$ for all possible values $x$ of $X$ and $y$ of $Y$.
That is indeed true if $X$ can't take the value $0$. But if $X = 0$, $XY = 0$ no matter what $Y$ is, so $\mathbb P(XY = 0\cdot y \mid X = 0) = 1 \ne \mathbb P(Y = y)$.