If $A\otimes_R S\cong \frac{S[x]}{(x^p)}$ then $A\cong \frac{R[y]}{(y^p-c)}$ where $R\to S$ is an fppf map

Question

This question generalizes my previous answered question.

Conjecture: Let $R$ be a commutative ring of characteristic $p$, $R\to S$ be an fppf ring map (i.e. faithfully flat and of finite presentation), $A$ be an $R$-algebra s.t. $A\otimes_R S\cong \frac{S[x_1,...,x_m]}{(x_1^p,...,x_m^p)}$, then $A\cong\frac{R[y_1,...,y_m]}{(y_1^p-c_1,...,y_m^p-c_m)}$ for some $c_i\in R$.

I can manage to show that $A$ is affine locally of the desired form, but I cannot glue them to give a global result. Here is the things I can show:

1. $A$ is free of rank $p^m$ and $\Omega^1_{A/R}$ is free of rank $m$ by fpqc descent, tag 05B2.
2. $A^p\subset R$: Abuse sheaf properties, it suffices to show that for all $\mathfrak{p}\in \mathop{\mathrm{Spec}}R$, $(A_\mathfrak{p})^p\subset R_\mathfrak{p}$. Fix $\mathfrak{p}$, we have $A_\mathfrak{p} \otimes_{R_\mathfrak{p}} S_\mathfrak{p}\cong \frac{S_\mathfrak{p}[x_i]}{(x_i^p)}$. We know $R_\mathfrak{p}\to S_\mathfrak{p}$ is faithfully flat by base change, in particular, $A_\mathfrak{p}$ embeds in $A_\mathfrak{p} \otimes_{R_\mathfrak{p}} S_\mathfrak{p}$ and $(A_\mathfrak{p})^p\in S_\mathfrak{p}\cap A_\mathfrak{p}$. As local rings are Hermite rings, $A_\mathfrak{p}$ has an $R_\mathfrak{p}$-basis containing 1, for details see my answer in this mathStackExchange question, so $S_\mathfrak{p}\cap A_\mathfrak{p}=R_\mathfrak{p}$ and the result follows.
3. For each $\mathfrak{p}\in \mathop{\mathrm{Spec}}R$, $S\otimes_R \kappa(\mathfrak{p})$ is a non-zero, finitely presented $\kappa(\mathfrak{p})$-algebra. In particular, one of its maximal ideal $\mathfrak{q}$ has a residue field $\kappa(\mathfrak{q})$ finite over $\kappa(\mathfrak{p})$.
4. $A\otimes_R \kappa(\mathfrak{p}) \otimes_{\kappa(\mathfrak{p})} \kappa(\mathfrak{q})\cong \frac{\kappa(\mathfrak{q})[x_i]}{(x_i^p)}$, using arguments from this answer we have $A\otimes_R \kappa(\mathfrak{p})\cong \frac{A_\mathfrak{p}}{\mathfrak{p}A_\mathfrak{p}}\cong \frac{R_\mathfrak{p}}{\mathfrak{p}R_\mathfrak{p}}[Y_i]/(Y_i^p-c_i)$ for some $c_i\in \frac{R_\mathfrak{p}}{\mathfrak{p}R_\mathfrak{p}}$. Lift $y_i$ (image of $Y_i$) to $A_\mathfrak{p}$, we know $A_\mathfrak{p}$ is generated as an $R_\mathfrak{p}$-module by $y_1^{\{0,...,p−1\}}\cdots y_m^{\{0,...,p−1\}}$ by Nakayama, tag 00DV. So it is a basis by tag 05G8. It follows that $A_\mathfrak{p}\cong \frac{R_\mathfrak{p}[Y_i]}{(Y_i^p-c_i)}$ (we always have $c_i=y_i^p$).
5. With some care, we can show that there exists $f\in R\backslash \mathfrak{p}$ s.t. $A_f\cong \frac{R_f[Y_i]}{(Y_i^p-c_i)}$ .

This is as far as I can get. Now it remains to show $A\cong \frac{R[y_i]}{(y_i^p-c_i)}$. Is it provable or is there a counter example (maybe with a disconnected base)?