Maxwell's equations in vacuum are symmetric bar the problem with units that you have identified. In SI units
$$ \nabla \cdot {\bf E} = 0\ \ \ \ \ \ \nabla \cdot {\bf B} =0$$
$$ \nabla \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}\ \ \ \ \ \ \nabla \times {\bf B} = \mu_0 \epsilon_0 \frac{\partial {\bf E}}{\partial t}$$
If we let $\mu_0=1$, $\epsilon_0 =1$ (effectively saying we are adopting a system of units where $c=1$, then these equations become completely symmetric to the exchange of ${\bf E}$ and ${\bf B}$ except for the minus sign in Faraday's law. They are symmetric to a rotation (see below).
If the source terms are introduced then this breaks the symmetry, but only because we apparently inhabit a universe where magnetic monopoles do not exist. If they did, then Maxwell's equations can be written symmetrically. We suppose a magnetic charge density $\rho_m$ and a magnetic current density ${\bf J_{m}}$, then we write
$$ \nabla \cdot {\bf E} = \rho\ \ \ \ \ \ \nabla \cdot {\bf B} = \rho_m$$
$$ \nabla \times {\bf E} = -\frac{\partial {\bf B}}{\partial t} - {\bf J_m}\ \ \ \ \ \ \nabla \times {\bf B} = \frac{\partial {\bf E}}{\partial t} + {\bf J}$$
With these definitions, Maxwell's equations acquire symmetry to duality transformations.
If you put $\rho$ and $\rho_m$; ${\bf J}$ and ${\bf J_m}$; ${\bf E}$ and ${\bf H}$; ${\bf D}$ and ${\bf B}$ into column matrices and operate on them all with a rotation matrix of the form
$$ \left( \begin{array}{cc} \cos \phi & -\sin \phi \\ \sin \phi & \cos \phi \end{array} \right),$$
where $\phi$ is some rotation angle, then the resulting transformed sources and fields also obey the same Maxwell's equations. For instance if $\phi=\pi/2$ then the E- and B-fields swap identities; electrons would have a magnetic charge, not an electric charge and so on.
Whilst one can argue then about what we define as electric and magnetic charges, it is an empirical fact at present that whatever the ratio of electric to magnetic charge (because any ratio can be made to satisfy the symmetric Maxwell's equations) all particles appear to have the same ratio, so we choose to fix it that one of the charge types is always zero - i.e. no magnetic monopoles.
I mention all this really as a curiosity. It seems to me that the real symmetries of Maxwell's equations only emerge when one considers the electromagnetic potentials.
e.g. if we insert $B = \nabla \times {\bf A}$ and $E= -{\bf \nabla V} - \partial {\bf A}/\partial t$ into our Ampere's law
$$\nabla \times (\nabla \times {\bf A}) = \frac{\partial}{\partial t} \left({\bf -\nabla V} - \frac{\partial {\bf A}}{\partial t}\right) +{\bf J}, $$
$$-\nabla^2 {\bf A} +\nabla(\nabla \cdot {\bf A}) = -\nabla \frac{\partial V}{\partial t} - \frac{\partial^2 {\bf A}}{\partial t^2} + {\bf J}.$$
Then using the Lorenz gauge
$$\nabla \cdot {\bf A} + \frac{\partial V}{\partial t} = 0$$
we can get
$$ \nabla^2 {\bf A} - \frac{\partial^2 {\bf A}}{\partial t^2} + {\bf J} = 0$$
A so-called inhomogeneous wave equation.
A similar set of operations on Gauss's law yields
$$ \nabla^2 V - \frac{\partial^2 V}{\partial t^2} + \rho= 0$$
These remarkably symmetric equations betray the close connection between relativity and electromagnetism and that electric and magnetic fields are actually part of the electromagnetic field. Whether one observes $\rho$ or ${\bf J}$; ${\bf E}$ or ${\bf B}$, is entirely dependent on frame of reference.
The electric field $E$ is the field we apply, what we express with the first Maxwell equation is that its sources must come from the total density charge $\rho$. In a material, there will be some fixed charges, so the presence of $E$ will induce some dipoles, and this will make Polarization $P$ appear. Then polarization is related with the bound charge density $\rho_b$, while the rest of charges that are free to move we associate it with free density charge $\rho_f$ In vacuum, there is no bound density charge, so we have $\rho = \rho_f$.
In a presence of a material, we define displacement field $D$, or response field as
$$
D= \epsilon_0 E + P
$$
We normally consider the ideal situation for linear, homogeneous and isotropic material such that $P=\chi \epsilon_0 E$, where $\chi$ is electric susceptibility. In that way we could write $D=\epsilon_r \epsilon_0 E$, where $\epsilon_r = 1+ \chi$ is relative electric permitivity. $D$ is then taking into account the presence of free charges and bounded charges, althoug its sources are only the free charges. This can be seen easily, because the contribution of the Polarization is actually negative. It can be shown that $\nabla \cdot P =-\rho_b$, and as we know $\nabla \cdot E = \frac{\rho}{\epsilon_0}$. So if we take the divergence of the $D$ defined above and we use this results we have:
$$
\nabla \cdot D = \epsilon_0 \nabla \cdot E + \nabla \cdot P = \rho - \rho_b =\rho_f
$$
since by definition $\rho = \rho_f + \rho_b$
Best Answer
These equations are wrong. Among other things, they do not obey charge conservation, since according to them $$ \begin{align*} \vec{\nabla} \cdot \left(\vec{\nabla} \times \vec{B}\right) &=-\frac{1}{c^2}\vec{\nabla} \cdot \left( \frac{\partial \vec{E}}{\partial t}\right) +\frac{1}{\epsilon_0 c^2} \vec{\nabla} \cdot \vec{J} \\ 0 &= -\frac{1}{c^2}\frac{\partial }{\partial t}\left(\vec{\nabla} \cdot \vec{E}\right) +\frac{1}{\epsilon_0 c^2} \vec{\nabla} \cdot \vec{J} \\ &= \frac{1}{\epsilon_0 c^2} \left( -\frac{\partial \rho}{\partial t} + \vec{\nabla} \cdot \vec{J} \right) \end{align*} $$ This would imply that $\partial \rho/\partial t = \vec{\nabla} \cdot \vec{J}$, which would mean (for example) that the charge density in a particular region of space would increase when a positive current flowed out of it.