The following was resolved through personal communication, but I thought I'd post it here in case anyone has a similar problem in the future.

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I'm going to assume $P$ is nonsingular (strictly positive) so that $P^{-1}$ exists since $P$ is bounded.

As seen above $Q$ must commute with every _positive_ operator. One can show that this implies $Q$ must commute with _every_ arbitrary operator on $\mathcal{H}$. Therefore we must require $Q=2aI$ where $a\geq 0$ since $Q$ is positive [[1]](https://math.stackexchange.com/questions/27808/a-linear-operator-commuting-with-all-such-operators-is-a-scalar-multiple-of-the).

Then if we let $X = cP^{-1}$, the solution to the Lyapunov equation given when $H=I$, we can see that
$$
    XPH + HPX + HQ = 0
$$
for all $H$. This explains why in my numerical experiments just solving when $H=I$ works in all cases.