Conjecture on the floor-function sum for two primes congruent to 3 modulo 4

Let $p$ and $q$ be distinct primes with $p\equiv q\equiv3\pmod{4}$, let $\alpha,\beta\geq1$, and define

$$S_n:=\frac{1}{n}\sum_{k=1}^{n-1}\operatorname{Rem}(k^2\div n).$$

Set $h^*(-p)=h(-p)$ for $p\geq7$ and $h^*(-3)=1/3$. The same-congruence class-number conjecture. For $n=p^\alpha q^\beta$,

$$S_n=\frac{n-p^{\lfloor\alpha/2\rfloor}q^{\lfloor\beta/2\rfloor}}{2}-\left(\frac{q^{\lfloor\beta/2\rfloor}-1}{q-1}\left(\frac pq\right)+\frac{q^{\lfloor\beta/2\rfloor+1}-1}{q-1}\right)\frac{p^{\lfloor(\alpha+1)/2\rfloor}-1}{p-1}h^*(-p)-\left(\frac{p^{\lfloor\alpha/2\rfloor}-1}{p-1}\left(\frac qp\right)+\frac{p^{\lfloor\alpha/2\rfloor+1}-1}{p-1}\right)\frac{q^{\lfloor(\beta+1)/2\rfloor}-1}{q-1}h^*(-q).$$

The case $\alpha=\beta=1$ specializes to the displayed formula for $S_{pq}$ and yields the corresponding formula for $f(pq)$; the general identity is presented as conjectural.