Let $R$ be a domain and $K$ be fractional field. Let $I$ an invertible $R$-ideal. Show that $I$ is proper, i.e. $I:I=R$. Deduce that an additive subgroup $I \subset K = Q(R)$ is an invertible ideal for at most one subring of $K$.
By using the property $A:BC=(A:B):C$ then we obtain $I:I=(R:I^{-1}):I=R:R$ but I can not show that $R:R=\{ x \in K : xR \subset R \}=R$.
Edit: As the comment, in particular, $x.1 \in R$, hence $x \in R$.