I assume you want $q$ and $r$ to be odd primes. Also, note that I will be using the notation that $\zeta_m$ means an arbitrary primitive $m$-th root of unity (but the same one every time it appears in an equation), and will be proving the statement in that generality.

Lemma: For any odd $m>1$ and any $\zeta_m$, the number $\zeta_m+1$ is a unit.

Proof: Let $r$ be such that $m | 2^r-1$. We'll abbreviate $\zeta_m$ to $\zeta$.

Then $\zeta^{2^r} = \zeta$ so $$1 = \left( \frac{\zeta^{2}-1}{\zeta -1} \right) \left( \frac{\zeta^{4}-1}{\zeta^{2} -1} \right) \cdots \left( \frac{\zeta^{2^r}-1}{\zeta^{2^{r-1}} -1} \right)=$$ $$\left( \zeta+1 \right) \left( \zeta^{2} + 1 \right) \cdots \left( \zeta^{2^{r-1}} +1 \right),$$ exhibiting an explicit inverse for $\zeta+1$.

Note that, for $m$ odd, the negation of a primitive $2m$-th root of unity is a primitive $m$-th root of unity. So the negation of your proposed unit is $\zeta_q+\zeta_{-q} + \zeta_{r} + \zeta_{-r}$. There is a primitive $qr$-th root of unity, $\eta$, such that $\zeta_q = \eta^r$ and $\zeta_r = \eta^q$. So the claim is that $$\eta^r + \eta^{-r} + \eta^q + \eta^{-q}$$ is a unit. This factors as $$\eta^r \left(1 + \eta^{q-r} +\eta^{-q-r} + \eta^{-2r} \right) = \eta^r \left( 1+ \eta^{q-r} \right) \left( 1+\eta^{-q-r} \right).$$ Using that $q$ and $r$ are primes, we have $GCD(q-r, qr) = GCD(q+r,qr)=1$, so $\eta^{q-r}$ and $\eta^{q+r}$ are primitive $qr$-th roots of unity, and we are done by the lemma.

Note: I don't think I used that $q$ and $r$ were odd primes, only that they were odd, relatively prime, and $>1$.

I assume you want $q$ and $r$ to be odd primes. Also, note that I will be using the notation that $\zeta_m$ means an arbitrary primitive $m$-th root of unity (but the same one every time it appears in an equation), and will be proving the statement in that generality.

Lemma: For any odd $m>1$ and any $\zeta_m$, the number $\zeta_m+1$ is a unit.

Proof: Let $r$ be such that $m | 2^r-1$. We'll abbreviate $\zeta_m$ to $\zeta$.

Then $\zeta^{2^r} = \zeta$ so $$1 = \left( \frac{\zeta^{2}-1}{\zeta -1} \right) \left( \frac{\zeta^{4}-1}{\zeta^{2} -1} \right) \cdots \left( \frac{\zeta^{2^r}-1}{\zeta^{2^{r-1}} -1} \right)=$$ $$\left( \zeta+1 \right) \left( \zeta^{2} + 1 \right) \cdots \left( \zeta^{2^{r-1}} +1 \right),$$ exhibiting an explicit inverse for $\zeta+1$.

Let $\eta$ be a primitive $2qr$ root of unity. Then your proposed unit is $\eta^{r}+\eta^{-r} + \eta^q + \eta^{-q}$ and factors as $$\eta^r (1+\eta^{q-r})(1+\eta^{-q-r}).$$ Since $q$ and $r$ are odd and relatively prime, $\eta^{q-r}$ and   $\eta^{q+r}$ are primitive $qr$-th roots of unity and we are done by the lemma.