The answer is also yes if one of the primes, say $r$, is $2$, because then $\zeta_{2r}+\zeta_{2r}^{-1}=0$ and $\zeta_{2q}+\zeta_{2q}^{-1}=\zeta_{2q}(1+\zeta_{2q}^{-2})$ is a unit (as $\zeta_{2q}^{-2}$ is a primitive $q$th root of $1$ and $q$ is an odd prime).

Edit: Note that if both primes are odd then $\zeta_q+\zeta_q^{-1}+\zeta_r+\zeta_r^{-1}$ is also a unit, because when you multiply all terms by $-1$ you get a $\zeta_{2q}+\zeta_{2q}^{-1}+\zeta_{2r}+\zeta_{2r}^{-1}$. But this conclusion is false if one of the primes is $2$.