 Solution

The intersection $J=I\cap \mathbb{Z}$ is an additive subgroup of $\mathbb{Z}$, and $nj\in I\cap \mathbb{Z}$ for $n\in \mathbb{Z}$, $j\in J$. Thus it is an ideal of $\mathbb{Z}$, necessarily $d\mathbb{Z}$ for a unique $d\ge 0$.

Because $d\in I$, one has $dR\subseteq I$. Conversely, if $a/b\in I$ with $b$ coprime to $30$, multiplication by $b\in R$ gives $a\in I\cap \mathbb{Z}=d\mathbb{Z}$. Write $a=dk$; then $a/b=d(k/b)\in dR$. Thus $I=dR$.

Every positive integer factors as $d=2^{\alpha }{3}^{\beta }{5}^{\gamma }u$ with $\gcd (u,30)=1$. The factor $u$ is a unit, so it does not change the ideal. The complete distinct list isThey are distinct because the intersection of the displayed nonzero ideal with $\mathbb{Z}$ is $2^{\alpha }{3}^{\beta }{5}^{\gamma }\mathbb{Z}$: admissible denominators cannot cancel any of these prime powers. The triple $(0,0,0)$ gives $R$ itself.