I can prove the following result without too much trouble:

Let $f: X \to S$ be a proper flat morphism of finite presentation with $S$ irreducible having generic point $\xi \in S$. Then the following are equivalent,

(1) the generic fiber $X_\xi \to \mathrm{Spec}{\kappa(\xi)}$ is smooth

(2) there exists a smooth fiber $X_s \to \mathrm{Spec}(\kappa(s))$ for some $s \in S$

(3) there exists a dense open set $U \subset S$ such that $f : X_U \to U$ is smooth.

A possible reference is [EGA-IV-4-12.2.4] then use properness to conclude that the image of the nonsmooth locus is closed.

However, I am interested in the case that $f$ is not proper. It is easy to show that (2) does not imply (3) without properness. E.g. take $$ X = \mathrm{Spec}{k[x, y, z]/(y(xz - 1))} \to \mathrm{Spec}{k[z]} = \mathbb{A}^1_k $$ which has only one smooth fiber (over $z = 0$). However, is it true that (1) still implies (3) without properness? Explicitly, is the following true:

Let $f: X \to S$ be a flat morphism of finite presentation with $S$ irreducible having generic point $\xi \in S$. Then if $X_\xi \to \mathrm{Spec}(\kappa(\xi))$ is smooth then there is a dense open set $U \subset S$ such that $f : X_U \to U$ is smooth?

If this is true can somebody point me to a reference? If it is false can somebody provide a counterexample?

Many thanks.