When does the inverse limit preserve the localisation?

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Question

When is the following true?
$$\varprojlim(S_\alpha^{-1}A_\alpha)\cong(\varprojlim S_\alpha)^{-1}(\varprojlim A_\alpha)$$
(For details, consult the next part.)

Notations One can consult this part only when needed.
Let $I$ be a poset, not necessarily directed, let $\{A_\alpha,\phi_{\beta\alpha}:A_\beta\rightarrow A_\alpha\}$ be an inverse system of rings, let $A=\varprojlim A_\alpha,$ and let $\phi_\alpha:A\rightarrow A_\alpha$ be the canonical homomorphism.
Let $S_\alpha$ be a multiplicative subset of $A_\alpha$ such that $\phi_{\beta\alpha}(S_\beta)\subset S_\alpha,$ so that the $S_\alpha$ form an inverse system of multiplicative sets, with $S=\varprojlim S_\alpha,$ and that $\phi_\alpha(S)\subset S_\alpha, \forall\alpha.$
For every $\alpha\le\beta,$ let $\rho_{\beta\alpha}$ be the unique homomorphism such that the following diagram commutes:
$$\begin{array}[c]{ccc} A_{\beta}&\stackrel{\phi_{\beta\alpha}}{\rightarrow}&A_{\alpha}\\ \downarrow&&\downarrow\\ S_{\beta}^{-1}A_{\beta}&\stackrel{\rho_{\beta\alpha}}{\rightarrow}&S_{\alpha}^{-1}A_{\alpha} \end{array}$$
Then $\{S_\alpha^{-1}A_{\alpha},\rho_{\beta\alpha}\}$ is an inverse system of rings; let $A'=\varprojlim S_\alpha^{-1}A_\alpha.$
Let $\psi_{\alpha}$ the unique homomorphism such that the following diagram commutes:
$$\begin{array}[c]{ccc} A&\stackrel{\phi_\alpha}{\rightarrow}&A_\alpha\\ \downarrow&&\downarrow\\ S^{-1}A&\stackrel{\psi_\alpha}{\rightarrow}&S_{\alpha}^{-1}A_\alpha \end{array}$$
Then $\psi_\alpha$ induce a homomorphism $\psi:S^{-1}A\rightarrow A'.$

And my question is: when is $\psi$ an isomorphism?

Of course this is different from the above simplified (but more general) question, but the isomorphism does not matter for me, see the motivation below.

Motivation
If $A$ is a Noetherian ring, I am able to show that, for any open subset $U$ of $X:=\operatorname{Spec}A,$ the ring $\Gamma(U,\mathscr O_X)=\varprojlim_{D(f)\subset U}A_f$ is a localisation of $A$ with respect to a multiplicative subset. However, when $A$ is arbitrary, I failed to show this, and I suspected that the above proposition is always true, while I could not find a way to prove it either.

P.S. I was wrong and the assertion is not true in general.
Thanks in advance for any answers or hints.