The displacement current density,
$$\mu_0 \epsilon_0 \frac{\partial \textbf{E}}{\partial t}$$
has units of (N/A^2) (C^2/Nm^2) N/Cs = Ns^2/(Cm^2)
which is not the same for current density:
C/sm^2
Where is the discrepancy?
[Physics] Units for displacement current density
electromagnetism
Related Solutions
It's because the divergence alone does not determine a vector field. You must know both the divergence and the curl to specify the field. (This is just math, no physics.)
Now the physics:
In general, curl($ \mathbf D $) = curl($\mathbf P$), so you can't simply think of it as an electric field due to the free charge.
In linear media, curl($ \mathbf{D}$)=0, so you in that case you can think of $\mathbf D$ as just the field due to free charges.
Basically, you are thinking if you can use intuition from the $\mathbf E$ field for the $\mathbf D$ field. But for the intuition part, you are most likely assuming electrostatic intuition, so that requires div($\mathbf E$)=$ \frac{\rho}{\epsilon_{0}} $ AND curl($\mathbf{E} $)=0.
Hence, for the $\mathbf D$ field you already know that div($\mathbf D$) = $\rho_{free}$ and curl($\mathbf D$) = curl($\mathbf P$). These equations will look "exactly" like the electrostatic div($\mathbf E$) =$\frac{\rho}{\epsilon_{0}}$, curl($\mathbf E$)=0 only if curl($\mathbf D$) vanishes. General speaking, it doesn't. But it does in linear media.
You did not make a mistake!
In Ampere's circuital law a line integral is evaluated $\displaystyle\oint_L \vec B \cdot d\vec l$ where $L$ is a closed loop.
That closed loop defines the edge of an open surface $S$ as shown in the diagram below which is an annotated picture of a butterfly net.
The important thing to realise is the there is no condition placed on the open surface $S$ other than the fact that the closed loop defines its edge.
If the closed loop $L$ does not move but the surface $S$ does move the value of the integral $\displaystyle\oint_L \vec B \cdot d\vec l$ must stay the same.
Imagine a wire leading carrying a current $I$ leading to the plate of a capacitor $C$.
For surface $S1$ there is no problem in equating the line integral $\displaystyle\oint_L \vec B \cdot d\vec l$ to $\mu_0I$ as is required by Ampere's law.
However the problem is surface $S2$ where the is no current passing through it and yet the line integral has a finite value unchanged from that when surface $S1$ was considered.
Now introduce a displacement current $\displaystyle \frac {d}{dt}\int_{S2} \vec D \cdot d\vec S$ which is has exactly the same value as $I$ the current passing through surface $S1$.
The problem goes away and there is agreement with your analysis in that the displacement current is independent of the material in the gap.
Best Answer
It's cause $\mu_0$ is not part of the displacement current density.