Let $g\in A$ be a regular element and $f\in A$, when do we have the preimage of $(x-f/g)\subset A_g[x]$ in $A[x]$ is $(gx-f)$?

Question

Let $A$ be a commutative ring with $1$. Let $g$ be a regular element of $A$, i.e. $g\neq 0$ and $g$ is not a zero-divisor of $A$. Let $A_g$ be the localization of $A$ at the set $\{g^n:n\geq 0\}$, then we have an injection $A\to A_g$.

Let $\{f_1,\dots,f_n\}\subset A$. Let $$J:=(x_1-f_1/g,\dots,x_n-f_n/g)\subset A_g[x_1,\dots,x_n]$$ $$I:=(gx_1-f_1,\dots,gx_n-f_n)\subset A[x_1,\dots,x_n]$$ Clearly we have $I\subset J\cap A[x_1,\dots,x_n]$.

I want to ask about the necessary and sufficient conditions on the triples $(A,g,\{f_1,\dots,f_n\})$ s.t. $I=J\cap A[x_1,\dots,x_n]$.

Some examples of necessary or sufficient conditions is as following:

1. If $g$ is a unit, then $A=A_g$ and we must have $I=J\cap A[x_1,\dots,x_n]=J$.
2. Say $n=1$ and $f_1=g\notin A^\times$, it's easy to see $x_1-1\in (x_1-1)\cap A[x_1]$ and $x_1-1\notin (gx_1-g)$.
3. Similarly if $n=1$ and $g\notin A^\times$ and there exists a non-unit divisor $s$ of $g$ s.t. $s$ divides $f_1$, then we have $\frac{g}{s} x-\frac{f_1}{s}\in (x_1-f_1/g)\cap A[x_1]$ and $\frac{g}{s} x-\frac{f_1}{s}\notin (gx_1-f_1)$.

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Update, I have found the necessay and sufficient condition for case $n=1$.

Lemma 1: The following are equivalent:

1. $\forall s\in A,g|fs\Rightarrow g|s$
2. $\forall m\geq1,\forall s\in A,g^{m}|fs\Rightarrow g^{m}|s$
3. $(gx-f)_{A[x]}=A[x]\cap(x-\frac{f}{g})_{A_{g}[x]}$

Proof: Clearly (2) implies (1). Assume (1) holds and (2) does not. There exists a pair $(m>1,s)$ s.t. $g^m|fs$ and $g^m\nmid s$. WLOG assume $m$ is minimal. Then $g^{m-1}|g^m|fs$, so $g^{m-1}|s$ by minimality of $m$. Say $s=g^{m-1}s^\prime$, clearly $g\nmid s^\prime$. Since $g^m|fs=fg^{m-1}s^{\prime}$, we have $g|fs^{\prime}$ and $g\nmid s^\prime$. It contradicts with (1).

Next we show that $(2)\Rightarrow(3)$. Note that $(x-\frac{f}{g})_{A_{g}[x]}=(gx-f)_{A_{g}[x]}$. Let $P(x)(gx-f)\in A[x]\cap(gx-f)_{A_{g}[x]}$ with $P(x)\in A_{g}[x]$. We want to show that $P(x)(gx-f)\in(gx-f)_{A[x]}$. Since $(gx-f)$ is a regular element of $A_{g}[x]$ (can be proved by induction), it is equivalent of showing that $P(x)\in A[x]$. Equivalently, we can show that all the coefficients of $P(x)$ are in $A$.

Let $P(x)$ be a minimal degree polynomial s.t. it doesn't hold. Let $\frac{s}{g^k}$ be its constant term. So the constant term of $P(x)(gx-f)\in A[x]$ is $\frac{fs}{g^{k}}\in A$. By (2) we have $\frac{s}{g^k}\in A$. It follows that $\frac{P(x)-s/g^k}{x}$ is another such polynomial with a lower degree, contradiction.

Next we show $(3)\Rightarrow (1)$. Assume (3) holds, let $s\in A$ s.t. $g|fs$. We have that $\frac{s}{g}(gx-f)=sx-\frac{fs}{g}\in A[x]\cap(gx-f)_{A_{g}[x]}=(gx-f)_{A[x]}$. It follows that $\frac{s}{g}\in A$. $\square$