If $A[[x]]$ is Noetherian, will $A$ be Noetherian?
Despite your claim, I think that $A=A[[x]]/(x)$. Take the map taking a general element $\sum_{n=0}^\infty a_nx^n \to a_0$. It is clearly a homomorphism from $A[[x]] \to A$. Then, the kernel contains every element with constant term zero. Isn't this equal to the ideal $(x)$? By the "quotients of noetherian rings are noetherian" argument you cite, this proves the result.
Let me know if there is a mistake in my logic.