Maybe this is an idiot question, but I've heard that $A[[x, y]] = A[[x]][[y]]$ does not hold for $A$ an arbitrary commutative ring with identity, so I would like to know a counterexample, since the obvious homomorphism seems to be an isomorphism because everything converges in the $I$-adic topology, so the structure of the ring seems to be preserved.
Thanks in advance.
Edit: My intention was to ask about the answer here saying that $A[[x, y]] = A[[x]][[y]]$ is not true for $\mathbf{topological}$ rings.
Edit: This answers the original version of the question, not the modified topological version.
This does hold for any commutative ring. Perhaps what you're thinking is that when we pass to Laurent series we have
$A((x,y)) \subsetneq A((x))((y))$.
On the right hand side we can have a single element in which all negative powers of $x$ appear, e.g. $\sum_{n \geq 0} x^{-n^2} y^n$.
On the left hand side for any fixed element there must be a least power of both $x$ and $y$.Curses, foiled again. Suppose $A$ is a field: then by $A((x,y))$ one usually means the fraction field of $A[[x,y]]$. This includes all doubly finite-tailed formal Laurent series but also a bit more: e.g.$\sum_{n \geq 0} x^{-n} y^n = \frac{x}{x-y}$.
This is very confusing. See e.g. here and here. In particular the comments to the answer given in the first link assert that the example I gave above of an explicit element of $A((x))(y)) \setminus A((x,y))$ is correct. I thought there was another MO answer giving the details of that, but I cannot find it at the moment.