Relativistic Maxwell
Since any differential operator is local (in close proximity), and (consequently) a finiteness of a propagation speed of the fields (in infinitely small distance) can not make adjustments to the equation. Therefore, Maxwell's equations in differential form in the context of the Master Theory would not be changed in comparison with the classics. Namely:
$$\vec{\nabla} \vec{E} = \frac{\rho}{\epsilon \epsilon_0} \, ; \; \vec{\nabla} \vec{B} = 0 \, ; \; \vec{\nabla} \times \vec{E} = - \frac{ \partial \vec{B} }{ \partial t } \, ; \; \vec{\nabla} \times \vec{B} = \mu \mu_0 \left( \vec{j} + \epsilon \epsilon_0 \frac{ \partial \vec{E} }{ \partial t } \right) \, ; \; \vec{E} = \sigma \vec{j}$$
But integral Maxwell (because of a nonlocality of a integral operators) will have a fundamentally different (non-classical, but - relativistic) form. This is explained by the fact that Stokes' and Gauss' theorems:
$$\oint _L \vec{A} d\vec{l} = \int _S \vec{\nabla} \times \vec{A} \; d\vec{s}$$
and
$$\int _S \vec{A} d\vec{s} = \int _V \vec{\nabla} \vec{A} dV$$
valid only for stationary fields, or fields, the speed of which is infinite. Application of these theorems (for the output of the integral form of Maxwell's equations) can not be correct in the relativistic case.
Corrections:
Your conventions for dot product and integrals are a little unclear.
Your Maxwell equations are for linear, isotropic matter where
$$\epsilon = 1 + \chi_e$$ is the relative permittivity and $$\mu = 1 + \chi_m$$ is the relative permeability. This differs from the common treatment where absolute permittivities and permeabilities are uses and differs from the microscopic fundamental theory is that a linear theory for matter is used rather than modeling the matter in EM theory itself. For example, there is no expectation of isotropism in moving matter.
So, to be clear, the currents and charges are
free charges and not the charges inside of atoms.
So $$\vec{\nabla} \cdot \vec{D} = \vec{\nabla} \cdot ( \epsilon_{\textrm{rel}} \epsilon_0 \vec{E} ) = \rho_{\textrm{free}}$$, $$\vec{\nabla} \cdot \vec{B} = 0$$, $$\vec{\nabla} \times \vec{E} = - \frac{ \partial \vec{B} }{ \partial t }$$, $$\vec{\nabla} \times \vec{H} = \vec{\nabla} \times ( \frac{1}{ \mu_{\textrm{rel}} \mu_0} \vec{B} ) = \vec{j}_{\textrm{free}} + \epsilon_{\textrm{rel}} \epsilon_0 \frac{ \partial \vec{E} }{ \partial t }$$ are not fundamental, but approximations to Maxwell's equations in isotropic, linear matter. Likewise, Ohm's law, is for isotropic, linear matter where time-varying currents are unimportant and is written as $$\vec{j}_{\textrm{free}} = \sigma \vec{E}$$ when $$\sigma$$ is conductivity.
When there are no charges bound to matter, we have Maxwell's equations in this form: $$\vec{\nabla} \cdot ( \epsilon_0 \vec{E} ) = \rho \, ; \; \vec{\nabla} \cdot \vec{B} = 0 \, ; \; \vec{\nabla} \times \vec{E} = - \frac{ \partial \vec{B} }{ \partial t } \, ; \; \vec{\nabla} \times ( \frac{1}{ \mu_0} \vec{B} ) = \vec{j} + \epsilon_0 \frac{ \partial \vec{E} }{ \partial t }$$
Instead of tracking 6 parameters throughout space and time, we can track 4 parameters if we write the above in terms of the scalar potential $$\varphi$$ and the vector potential, $$\vec{A}$$.
Then we define $$\vec{E} = - \vec{\nabla} \varphi - \frac{ \partial \vec{A} }{ \partial t }$$ and $$\vec{B} = \vec{\nabla} \times \vec{A}$$. Thus by definition, $$\vec{\nabla} \cdot \vec{B} = \vec{\nabla} \cdot \vec{\nabla} \times \vec{A} = 0$$ and $$\vec{\nabla} \times \vec{E} = - \vec{\nabla} \times \vec{\nabla} \varphi - \vec{\nabla} \times \frac{ \partial \vec{A} }{ \partial t } = 0 - \frac{ \partial \vec{\nabla} \times \vec{A} }{ \partial t } = - \frac{ \partial \vec{B} }{ \partial t }$$. So the physics content of Maxwell's equations reduce to these two:
$$\vec{\nabla} \cdot \vec{E} = - \vec{\nabla} \cdot \vec{\nabla} \varphi - \vec{\nabla} \cdot \frac{ \partial \vec{A} }{ \partial t } = - \nabla^2 \varphi - \frac{ \partial \vec{\nabla} \cdot \vec{A} }{ \partial t }= \frac{\rho}{\epsilon_0} $$ and $$\vec{\nabla} \times \vec{B} - \epsilon_0 \mu_0 \frac{ \partial \vec{E} }{ \partial t } = \vec{\nabla} \times \vec{\nabla} \times \vec{A} + \frac{1}{c^2} \frac{ \partial }{ \partial t } \left( \vec{\nabla} \varphi + \frac{ \partial \vec{A} }{ \partial t } \right) = \vec{\nabla} \vec{\nabla} \cdot \vec{A} - \nabla^2 \vec{A} +\vec{\nabla} \frac{1}{c^2} \frac{ \partial \varphi }{ \partial t } + \frac{1}{c^2} \frac{ \partial^2 \vec{A} }{ \partial t^2 } = \vec{\nabla} \left( \vec{\nabla} \cdot \vec{A} + \frac{1}{c^2} \frac{ \partial \varphi }{ \partial t } \right) + \frac{1}{c^2} \frac{ \partial^2 \vec{A} }{ \partial t^2 } - \nabla^2 \vec{A} = \mu_0 \vec{j}$$.
Although we have less variables, by using the potentials we introduce a choice of gauge which does not affect the physics. So I may freely choose the Lorentz Gauge: $$ \vec{\nabla} \cdot \vec{A} + \frac{1}{c^2} \frac{ \partial \varphi }{ \partial t } = 0$$.
So we have, $$\frac{1}{c^2} \frac{ \partial^2 \vec{A} }{ \partial t^2 } - \nabla^2 \vec{A} = \mu_0 \vec{j}$$ and $$- \nabla^2 \varphi - \frac{ \partial \vec{\nabla} \cdot \vec{A} }{ \partial t } = - \nabla^2 \varphi + \frac{1}{c^2} \frac{ \partial^2 \varphi }{ \partial t^2 } = \frac{\rho}{\epsilon_0}$$
Finally, we can go a step further if we work in four dimensions, not three. Then we have a single equation in four-vectors:
$$ \partial^2 A^{k} = \left( \frac{1}{c^2} \frac{ \partial^2 }{ \partial t^2 } - \nabla^2 \right) \begin{pmatrix} \frac{1}{c} \varphi \\ \vec{A} \end{pmatrix} = \mu_0 \begin{pmatrix} c \rho \\ \vec{j} \end{pmatrix} = \mu_0 J^{k} $$
Furthermore, this last expression begins to suggest the concept of Lorentz covariance rendering the laws of physics invariant.
All of this is carried in a standard electromagnetism course and all of it is available in Wikipedia pages. I have labored to make the connections explicit.
