You are right about $S(\alpha)\cap S(\beta)=\varnothing\iff \dfrac{n}{\alpha}+\dfrac{m}{\beta}=1$ for some $n,m\in\mathbb Z^+$ (see Theorem 8 of the cited paper).

The implication $\Longleftarrow$ is easy so let's assume that $\frac{n}{\alpha}+\frac{m}{\beta}\neq 1$ for all $n,m\in\mathbb Z^+$. Then one of the following is true.

- The numbers $1,\dfrac{1}{\alpha},\dfrac{1}{\beta}$ are linearly independent over $\mathbb Q$,
- there exist some $n,m,k\in\mathbb Z^+$ such that $\left|\dfrac{n}{\alpha}-\dfrac{m}{\beta}\right|=k$,
- there exist some $n,m,k\in\mathbb Z^+$ such that $\dfrac{n}{\alpha}+\dfrac{m}{\beta}=k$ with $k>1$, and $\gcd(n,m,k)=1$.

It is enough to show that any one of 1., 2. or 3. implies that $S(\alpha)\cap S(\beta)=\emptyset$. Elementary proofs of these can be found in this paper (Theorem 5, Theorem 6 and Theorem 7 respectively).