The assertion is false. For $n\geq2$, take the affine hypersurface
$$
X_n=\operatorname{Spec}k[x,y,z]/(xy-z^n)\subset\mathbb A_k^3.
$$
It is an <integral scheme>, and its only possible singular point is the origin, which has codimension two. Hence it is <regular in codimension one>; as a hypersurface it satisfies Serre's condition $(S_2)$, so the <Serre criterion for normality> also makes it a <normal scheme>. The <Divisor class group of an A-type surface singularity> is
$$
\operatorname{Cl}(X_n)\cong\mathbb Z/n\mathbb Z,
$$
generated by $V(x,z)$. Thus a closed affine subscheme satisfying $(\star)$ can have nonzero torsion in its class group.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 113 / 2 / c / Solution

{c}
{title2=$\operatorname{Cl}(k[x,y,z]/(xy-z^n))$}

For $n\geq2$, the normal affine surface
$$
X_n=\operatorname{Spec}k[x,y,z]/(xy-z^n)
$$
has $\operatorname{Cl}(X_n)\cong\mathbb Z/n\mathbb Z$. The class of the prime divisor $D=V(x,z)$ generates: localization at $x$ is a unique factorization domain, so the divisor-class localization sequence leaves only $D$, while $\operatorname{div}(x)=nD$ gives its exact order.


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