Some trivial observations. We have
$$P_{QR}(1/x) x^{(p-1)/2} = \prod_{QR} (1 - x \zeta^k),$$
$$P_{QNR}(1/x) x^{(p-1)/2} = \prod_{QNR} (1 - x \zeta^k),$$
which are easier to work with. On the other hand,
if $p \ge 5$, the product of $\zeta^k_p$ over quadratic residues is one,
and the product over non-residues is also one. Hence we can write
$$\begin{aligned}
P_{QR}(1/x) x^{(p-1)/2} = & \ \prod_{QR} ( 1 - x \zeta^k) = \prod_{QR} ( \zeta^{-k} - x) \\
= & \
(-1)^{(p-1)/2} \prod_{QR} (x - \zeta^{-k}) \\
= & \ \begin{cases} P_{QR}(x), & p \equiv 1 \mod 4 \\
- P_{QNR}(x), & p \equiv 3 \mod 4 \end{cases} \end{aligned}$$
because $(-1/p) = (-1)^{(p-1)/2}$. The same swapping occurs for $P_{QNR}$.
Let's consider the highest powers of $P_{QR}(x)$ and $P_{QNR}(x)$,
or equivalently the lowest powers of $P_{QR}(1/x) x^{(p-1)/2}$ and
$P_{QNR}(1/x) x^{(p-1)/2}$. We have
$$\frac{P_{QR}(1/x) x^{(p-1)/2}}{P_{QNR}(1/x) x^{(p-1)/2}}
= \prod ( 1 - x \zeta^k)^{ \left( \frac{k}{p}\right)}.$$
As mentioned there is the Gauss sum:
$$\sum \left( \frac{k}{p}\right) \zeta^{k} = \sqrt{p^*},$$
and similarly
$$\sum \left( \frac{k}{p}\right) \zeta^{nk} = \left( \frac{n}{p} \right) \sqrt{p^*},$$
applying $[n] \in (\mathbb{Z}/p\mathbb{Z})^{\times} = \mathrm{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$ to both sides; when $p|n$ we interpret the RHS as being zero, and this is still correct, although we won't actually care about terms this deep into the power series above.
We deduce that
$$- \log \left( \frac{P_{QR}(1/x) x^{(p-1)/2}}{P_{QNR}(1/x) x^{(p-1)/2}} \right) = \sum_{n=1}^{\infty} \frac{x^n}{n} \sum \left( \frac{k}{p}\right) \zeta^{nk}
= \sqrt{p^*} \sum_{n=1}^{\infty} \left( \frac{n}{p}\right) \frac{x^n}{n}.$$
Naturally
$$- \log \left( P_{QR}(1/x) x^{(p-1)/2} P_{QNR}(1/x) x^{(p-1)/2} \right)
= \log \left( \frac{1 - x^p}{1 - x} \right) = \log(1-x) + O(x^p) = - \sum \frac{x^n}{n},$$
and so (for example)
$$\log(P_{QR}(1/x) x^{(p-1)/2}) = \frac{1}{2} \sum \frac{x^n}{n} \left(1 - \left( \frac{n}{p}\right) \sqrt{p^*}\right) + O(x^p),$$
You can now formally expand this out to get the first few terms.
For example, the first non-zero term is
$$\frac{1 - \sqrt{p^*}}{2},$$
and the second is
$$ \frac{3 + p^* - 2 \sqrt{p^*}\left(1 + \left( \frac{2}{p}\right) \right) }{8}$$
For example, if $p \equiv 3,5 \mod 8$ so $(2/p) = -1$, this is
$$\frac{3 + p^*}{8}.$$
Note the conditions on $p$ ensure that this is an algebraic integer, as it has to be.
As you keep going, you get more and more terms involving the quadratic residues $(n/p)$ for small $n$,
and it becomes messier and messier, and dependent on $p$ modulo higher integers.
The third term, for example, is
$$\frac{15 - 9 \sqrt{p^*} + 3 p^* - p^* \sqrt{p^*} + 6(p^* - \sqrt{p^*})(2/p) - 8 \sqrt{p^*} (3/p)}{48}.$$
