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.

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].

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.

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.

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].

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.