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).
(1) Note that if both primes are odd then $\zeta_q+\zeta_q^{-1}+\zeta_r+\zeta_r^{-1}$ is also a unit. Indeed, because when you multiply all terms by $-1$ you get -\zeta_q$is a primitive$\zeta_{2q}+\zeta_{2q}^{-1}+\zeta_{2r}+\zeta_{2r}^{-1}$. But 2q$th root of $1$ (this conclusion relies on $q$ being odd), so let's call it $\zeta_{2q}$, and likewise for $r$. Then $-(\zeta_q+\zeta_q^{-1}+\zeta_r+\zeta_r^{-1})=\zeta_{2q}+\zeta_{2q}^{-1}+\zeta_{2r}+\zeta_{2r}^{-1}$, and we know that the latter is false a unit.
(2) Note also that if one of the primes, say $r$, is $2$.2$then$\zeta_q+\zeta_q^{-1}+\zeta_r+\zeta_r^{-1}$is not a unit: it is equal to$\zeta_q+\zeta_q^{-1}-2$, so it is a unit times the square of$\zeta_q-1$, but the latter (unlike$\zeta_q +1$) is not a unit because it goes to$0$under the unique ring homomorphism$\mathbb Z [\zeta_q]\to \mathbb Z/q$, which takes$\zeta_q$to$1$. 2 added 276 characters in body 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$2$. 1 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).