We have $\alpha\gamma=0^*\cup\{0\}\cup\{st:0t(1/r\notin\alpha)\}$. For positive $t$ the condition that there is some $r>t$ such that $1/r\notin\alpha$ is equivalent to the condition that there is a positive $r<1/t$ such that $r\notin\alpha$, i.e., that $\alpha\subsetneqq(1/t)^*$.
Let $q\in 1^*$; clearly $q\in\alpha\gamma$ if $q\le 0$, so assume that $0. Suppose that $q\notin\alpha\gamma$. Then for every positive $s\in\alpha$ and $t$ such that $\alpha\subsetneqq(1/t)^*$, $q\ne st$. Equivalently, for every positive $s\in\alpha$, $\alpha$ is not a proper subset of $(s/q)^*$, i.e., $(s/q)^*\subseteq\alpha$. Fix a positive $s_0\in\alpha$. Given $s_n$ for some $n\in\omega$, let $s_{n+1}=s_n/q$; an easy induction shows that ${s_n}^*\subseteq\alpha$ for each $n\in\omega$. But $s_n=s_0q^{-n}$, so you’ll have the desired contradiction showing that $q\in\alpha\gamma$ once you show that the sequence $\langle q^{-n}:n\in\omega\rangle$ is unbounded in $\Bbb Q$.
We know that $q=a/b$ for some positive integers $a$ and $b$ such that $a, so $q^{-n}=\left(\frac{b}a\right)^n=\left(1+\frac{b-a}a\right)^n\;.$
Let $p=\dfrac{b-a}a$; then $q^{-n}=(1+p)^n$, and it’s an easy induction to show that $(1+p)^n\ge 1+np$. Since $\langle 1+np:n\in\omega\rangle$ is clearly unbounded, we’re done.