@Smiley1000 : I don't think that works either. $|F'|/F^2=2|x|\cos(\exp(x^2))+o(x)$ doesn't tend to zero as $x\to\infty$, which proves $1/F$ is not Schwartz. But is $F$ reverse Schwartz? The denominator of $x^n/F'$ might surely cause problems (like having zeros after every $M>>0$).

@Smiley1000 : That's not reverse Schwartz according to the definition in the OP. For example, your $F$ is not positive: it has zeros for $x>M$ for every arbitrarily large $M$. This prevents $1/F\to0$.

Thank you very much, especially for the very fascinating detailed description of your thought process!