If special fiber of $S$ is smooth, can we say $S$ itself is regular?

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Let $A$ be a DVR with maximal ideal $P$ and $S$ the scheme over $A$ defined by $$S=\operatorname{Spec} A[X,Y,Z]/(Y^2Z-X^3-XZ^2).$$ If the special fiber of $S$ is smooth, can we say $S$ is regular ?

Here, $S$ is regular at $P$ if only if its local ring at $P$ is regular as a ring. And special fiver of $S$ is smooth if only if rank of its Jacobian(In this case $(-3X^2-Z^2, 2YZ, Y^2-2XZ)$) is nonzero. I think this is true in general scheme over DVR, but I have never heard such a claim.

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Let $f:S\rightarrow\mathrm{Spec}(A)$ be a closed morphism of finite type, where $A$ is a discrete valuation domain and $S$ is integral. Then $S$ is a regular scheme if its closed fibre $\overline{S}:=S\times_A k$, $k$ the residue field of $A$, is a regular scheme.

Proof: $\mathrm{dim}(S)=\mathrm{dim}(S\times_A k)+1$ - see for example Stacks Lemma 29.52.4.

The closedness of $f$ yields that the closure $\overline{\{y\}}$ for every point $y\in S$ contains a closed point $x\in\overline{S}$. Since the local ring $O_{S,y}$ then is a localization of the local ring $O_{S,x}$ and since regularity is stable under localization it suffices to prove the regularity of the local rings $O_{S,x}$ of points $x$ closed on $\overline{S}$.

For a prime element $t\in A$ one has $O_{\overline{S},x}=O_{S,x}/tO_{S,x}$, hence for the maximal ideal $M_{S,x}$ one gets $M_{S,x}=(t,a_1,\ldots ,a_d)$ for any set of elements $a_1,\ldots ,a_d\in M_{S,x}$ such that $a_1+tO_{S,x},\ldots ,a_d+tO_{S,x}$ generate $M_{\overline{S},x}$. Hence if one choses the later to be a regular system of parameters of $O_{\overline{S},x}$, then $t,a_1,\ldots ,a_d$ is a regular system of parameters of $O_{S,x}$.