Actually, the commutator relation $\left[a, a^†\right] = 1$ you cited does survive conversion, intact. The real issue is that the full conversion to the Heisenberg Picture does not preserve products, because neither the operator $𝔏$ nor its adjoint $𝔏^†$ satisfy the product (a.k.a. Leibnitz) rule.
You worked out the commutator wrong ... precisely because products are not preserved by the full conversion to the Heisenberg Picture. When you do it right, then after all the smoke clears, the commutator pops back up intact.
I'll explain "full" in a moment after expanding your example to include a unitary part:
$$\frac{dρ}{dt} = \frac{\left[Ea^†a, ρ\right]}{iħ} + 𝔏 ρ$$
with $E = ħω$ and with your
$$𝔏 ρ = κ\left(aρa^† - \frac{\left\{a^†a, ρ\right\}}2\right).$$
So, we added a Hamiltonian $H = Ea^†a$, for some constant $E$, with the understanding that your example can be recovered as the special case where $E = 0$.
The conversion between pictures takes place by matching the corresponding expectation values, which we'll denote in short-hand by
$$\left<Aρ\right> = \text{Tr}(Aρ),$$
for the expectation value of an observable $A$ in a state $ρ$. We will make use of the "cyclicity" property of the trace, as well, e.g.
$$\left<ρBA\right> = \left<AρB\right> = \left<BAρ\right>.$$
We will also make use of its linearity and factorability with respect to constants $k$:
$$\left<A ± B\right> = \left<A\right> ± \left<B\right>,\quad \left<kA\right> = k\left<A\right>.$$
Three pictures will be considered, here:
- The Schrödinger Picture: observables $A$ constant and eternal, states $ρ(t)$ "flow" in time $t$.
- The "Heisenberg" Picture: observables $A_H(t)$ are functions of time $t$ (and other coordinates, if dealing with field theory), states $ρ_H$ are eternal and denote an entire history of a system, i.e. its "time-line"; rather than an unfolding of a system over time, effectively treating $t$ as "block time", more akin to how Relativity treats it.
- The Lindblad Picture: observables $A_L(t)$ are functions of time $t$, as above that evolve unitarily, states $ρ_L(t)$ are also functions of time, but which evolve non-unitarily, as described below. In the limit of zero non-unitariness (here: as $κ → 0$), this becomes the Heisenberg Picture.
In the Lindblad Picture, the evolution of $A_L$ is given only by the unitary part of the Lindblad equation - the part involving $H$:
$$\frac{dA_L}{dt} = \frac{\left[A_L, H\right]}{iħ}.$$
The key equations that characterize the other two pictures are
$$\frac{dA}{dt} = 0,\quad \frac{dρ_H}{dt} = 0,$$
i.e. that Schrödinger Picture observables $A$ are "eternal" in the sense that they don't flow in time, and that "Heisenberg" Picture states $ρ_H$ are "eternal", and represent an entire history of a system, rather than its state flowing in time.
For convenience, we will also set $ρ_H = ρ(0) = ρ_L(0)$ and $A_H(0) = A = A_L(0)$, to tie all three forms of observables and states to each other. This ties two sets of evolutions to a specific time $t = 0$, which could be treated as a distinguished "now". Choosing a different $t = t_0$ will lead to altered forms of the transforms, so that they are "now"-dependent.
Because the operator on the right-hand side of the equation for $A_L$ satisfies the product (i.e. Leibnitz) rule, then conversions of operators to the Lindblad Picture will preserve products, i.e. $(AB)_L = A_L B_L$. For your example, it will turn out that $H = H_L$, so we won't have to make any distinction between the different forms of $H$.
Such will not be the case with the so-called "Heisenberg" picture because the conversion made was "full" in the sense that the Lindblad operators were also moved to the evolution equation for the observables, and not just the Hamiltonian $H$, so that, in general, $(AB)_H ≠ A_H B_H$. The mixing together of both the unitary part ($H$) and non-unitary part ($𝔏$) into the evolution equation for observables is actually the source of the incompatibility that your question points to. Instead, the conversion will only satisfy linearity and factorability $(A + B)_H = A_H + B_H$ and $(kA)_H = k A_H$ for constant $k$.
In this sense, the Lindblad Picture is better regarded as the analogue of the Heisenberg Picture for non-unitary dynamics, than is the "Heisenberg" Picture, which is why I put the scare quotes on the name. It repairs the product incompatibility.
The correspondence between the different pictures is given by
$$\left<Aρ(t)\right> = \left<A_L(t)ρ_L(t)\right> = \left<A_H(t)ρ_H\right>.$$
From this, the relevant evolution equations can be derived. First, we pick an arbitrary state $ρ_H$
$$\left<\frac{dA_H}{dt}ρ_H\right> = \frac{d}{dt}\left<A_Hρ_H\right> = \left<A\frac{dρ}{dt}\right>.$$
Then, we apply the (modified) evolution equation for $ρ$ and use linearity and factorability:
$$
\left<A\frac{dρ}{dt}\right> = \frac{\left<A\left[H, ρ\right]\right>}{iħ} + κ\left(\left<Aaρa^†\right> - \frac{\left<A\left\{a^†a, ρ\right\}\right>}2\right).
$$
After applying linearity and cyclicity:
$$\begin{align}
\left<A\left[H, ρ\right]\right> &= \left<A(Hρ - ρH)\right>\\
&= \left<AHρ\right> - \left<AρH\right>\\
&= \left<AHρ\right> - \left<HAρ\right>\\
&= \left<(AH - HA)ρ\right> = \left<\left[A,H\right]ρ\right>,\\
\left<Aaρa^†\right> &= \left<a^†Aaρ\right>,\\
\left<A\left\{a^†a, ρ\right\}\right> &= \left<A(a^†aρ + ρa^†a)\right>\\
&= \left<Aa^†aρ\right> + \left<Aρa^†a\right>\\
&= \left<Aa^†aρ\right> + \left<a^†aAρ\right>\\
&= \left<(Aa^†a + a^†aA)ρ\right> = \left<\left\{A,a^†a\right\}ρ\right>,
\end{align}$$
the result is:
$$\begin{align}
\left<A\frac{dρ}{dt}\right> &= \frac{\left<\left[A,H\right]ρ\right>}{iħ} + κ\left(\left<a^†Aaρ\right> - \frac{\left<\left\{A,a^†a\right\}ρ\right>}2\right)\\
&= \left<\left(\frac{\left[A,H\right]}{iħ} + κ\left(a^†Aa - \frac{\left\{A,a^†a\right\}}2\right)\right)ρ\right>\\
&= \left<\left(\frac{\left[A,H\right]}{iħ} + κ\left(a^†Aa - \frac{\left\{A,a^†a\right\}}2\right)\right)_Hρ_H\right>.
\end{align}$$
Applying linearity and factorability, then after noting the arbitrariness of the choice of $ρ_H$, we can extract out the observable part of this equation and write:
$$
\frac{dA_H}{dt} = \frac{\left[A,H\right]_H}{iħ} + \left(𝔏^†A\right)_H,
$$
which introduces the adjoint:
$$𝔏^†A = κ\left(a^†Aa - \frac{\left\{A,a^†a\right\}}2\right).$$
So, now we can see what this does to the various combinations $a$, $a^†$, $aa^†$ and $a^†a$ and what they each look like after conversion. First, as far as the commutator relation $\left[a,a^†\right] = 1$ goes, there's no issue, as you can see here:
$$\left<\left[a,a^†\right]_Hρ_H\right> = \left<\left[a,a^†\right]ρ\right> = \left<ρ\right> = \left<ρ_H\right>.$$
Thus, $\left[a,a^†\right]_H = 1$. Therefore, it also follows that $(aa^†)_H - (a^†a)_H = 1$.
To find out what the conversions of the different combinations are, first apply $H = Ea^†a$ and $𝔏^†$ to each of them:
$$
\left[H,a\right] = -Ea,\quad 𝔏^†a = -\frac{κ}{2}a,\\
\left[H,a^†\right] = +Ea^†,\quad 𝔏^†a^† = -\frac{κ}{2}a^†,\\
\left[H,aa^†\right] = 0,\quad 𝔏^†aa^† = -κa^†a = -κaa^† + κ,\\
\left[H,a^†a\right] = 0,\quad 𝔏^†a^†a = -κa^†a.
$$
The corresponding differential equations (along with the "now at time 0" initial conditions), are:
$$
\frac{da_H}{dt} = \frac{-Ea_H}{iħ} - \frac{κ}{2}a_H,\quad a_H(0) = a,\\
\frac{d(a^†)_H}{dt} = \frac{+E(a^†)_H}{iħ} - \frac{κ}{2}(a^†)_H,\quad (a^†)_H(0) = a^†,\\
\frac{d(aa^†)_H}{dt} = -κ(aa^†)_H + κ,\quad (aa^†)_H(0) = aa^†,\\
\frac{d(a^†a)_H}{dt} = -κ(a^†a)_H,\quad (a^†a)_H(0) = a^†a.\\
$$
The solutions, using $E = ħω$, are
$$
a_H = a e^{+iωt} e^{-κt/2},\quad
(a^†)_H = a^† e^{-iωt} e^{-κt/2},\quad
(aa^†)_H = aa^† e^{-κt} + 1 - e^{-κt},\quad
(a^†a)_H = a^†a e^{-κt}
$$
The commutator relation can be confirmed from this:
$$\begin{align}
\left[a,a^†\right]_H &= (aa^†)_H - (a^†a)_H\\
&= (aa^† - a^†a) e^{-κt} + 1 - e^{-κt}\\
&= 1 + (\left[a,a^†\right] - 1) e^{-κt}\\
&= 1.
\end{align}$$
The "product defects", arising from the non-Leibnitz nature of $𝔏^†$, can also be worked out:
$$(aa^†)_H - a_H(a^†)_H = 1 - e^{-κt},\quad (a^†a)_H - (a^†)_Ha_H = 0.$$
Finally, we have your version of the commutator - which is the incorrect version:
$$\left[a_H, (a^†)_H\right] = e^{-κt} ≠ 1 = \left[a, a^†\right]_H.$$
As noted above, regarding the "now at time 0" condition: if we choose a different time $t = t_0$ other than $t = 0$ to match up the different pictures, the conversions will be altered. This can be thought of as shifting from one "now" to a different "now" $t_0$. So, there is a concept of "now" built into these constructions, not just time flow. The "now" point represents a point of tangency where flow time meets block time.
For the Lindblad Picture, the resulting equation may be obtained by setting $κ = 0$:
$$
\frac{dρ_L}{dt} = (𝔏ρ)_L = κ\left(a^†ρa - \frac{\left\{aa^†,ρ\right\}}2\right)_L = κ\left((a^†)_Lρ_La_L - \frac{\left\{a_L(a^†)_L,ρ_L\right\}}2\right).
$$
The conversions for $a$, $a^†$, $aa^†$ and $a^†a$ are just
$$
a_L = a e^{+iωt},\quad
(a^†)_L = a^† e^{-iωt},\quad
(aa^†)_L = aa^† = a_L (a^†)_L,\quad
(a^†a)_L = a^†a = (a^†)_L a_L.
$$
So, the differential equation for observables actually simplifies further to:
$$\frac{dρ_L}{dt} = κ\left(a^†ρ_La - \frac{\left\{aa^†,ρ_L\right\}}2\right).$$
Shifting to a different "now" $t_0$ will add in phase factors $a_L → a_L e^{-iωt_0}$ and $(a^†)_L → (a^†)_L e^{+iωt_0}$ and effectively just time-shift the Lindblad states $ρ_L$, since the differential equation is symmetric under time translation.
The nuanced distinction between the Schrödinger, Lindblad and Heisenberg Pictures is masked by your example, because you only had a non-unitary part for $𝔏$ to your time evolution equation and no unitary part for $H$.
Further elaboration on the discussion here and a more generalized treatment, can be found in the subsequent reply.
