The statement comes from wiki, see the picture below.
I know that if $X$ is étale over $S$ then $F_{X/S}$ is an isomorphism and a proof can be found here. But is the reverse true? I believe that I have found a counter example.
Let $R$ be a domain of characteristic $p$ and $F$ be its fraction field. Denote the Frobenius map $F_R:R\to R'$ which is injective for a domain. Denote $S=R-0$. Then $$F\otimes_{R} R'=(S^{-1}R)\otimes_R R' =(F_R (S))^{-1}R'=(S^p)^{-1}R'$$ But $(S^p)^{-1}R'=S^{-1}R'=F'$ since $\frac{1}{t}=\frac{t^{p-1}}{t^p}$. So $F\otimes_R R'=F'$. Explicitly the relative Frobenius map $F_{F/R}:F\otimes_R R'\to F',\frac{u}{v}\otimes t\mapsto \frac{u^p t}{v^p}$ has an inverse $\phi:F'\to F\otimes_R R',\frac{x}{y}\mapsto \frac{1}{y}\otimes y^{p-1}x$. Now we just need to show there exists a characteristic $p$ domain $R$ s.t. $R\hookrightarrow F$ is not étale.
Let $K$ be a field of characteristic $p$, and we consider the localization $R=K[x]\hookrightarrow K(x)=F$ which is not finitely generated (if so then $K(x)$ is finitely generated over $K$, contradiction), hence not étale.