Definitions first:

Let $R$ be a ring with unity, and $\sigma$ an endomorphism of $R$. The (left) Ore extension $R[x;\sigma]$ is the free left $R$-module with basis $\{1,x,x^2, \dots\}$, made into a ring by the multiplication $xr = \sigma(r)x$ for every $r \in R$. Intuitively, it's a non-commutative version of a polynomial ring.

The skew power series ring $R[[x;\sigma]]$ is defined in the same way but now infinite sums of powers of $x$ are allowed (but still no negative powers).

The following is an approach towards the answer but I'm a bit unsure of the details towards the end.

So let's take our two rings $R[x;\sigma]$ and $S[[y;\sigma']]$ and suppose there exists an isomorphism $\varphi: R[x;\sigma] \rightarrow S[[y;\sigma']]$.

We have $xr = \sigma(r)x$ and $ys = \sigma'(s)y$ for all $r \in R$, $s \in S$, and the isomorphism has to respect these identities.

Set $a = \varphi^{-1}(y)$. Now for any $r \in R$, we can look at how $r$ and $a$ interact:

\[ar = \varphi^{-1}(ys) = \varphi^{-1}(\sigma'(s)y) = \varphi^{-1}(\sigma'(s))a = \varphi^{-1}(\sigma'(\varphi(r)))a\]
where $\varphi^{-1}(r) = s \in S[y;\sigma']$.

Let $\psi = \varphi^{-1}\circ \sigma' \circ \varphi$, which is an endomorphism of $R$. So $R[a;\psi]$ is also a skew polynomial ring, and this should? be isomorphic to $R[x;\sigma]$ (Maybe we need to assume $y$ is not a zero-divisor?)

Now we have $R[a;\psi] \cong R[x;\sigma] \cong S[[y;\sigma']]$; composing gives us an isomorphism $R[a;\psi] \rightarrow S[[y;\sigma']]$ which sends $a$ to $y$, which seems unlikely.

(Sorry, quite a bit of hand-waving towards the end there! You can also show that $1+ax$ must be invertible in $R[x;\sigma]$, similarly to the post linked to in the question, and then maybe use the existence of a commuting rule for $a$ to produce a contradiction?)