Well, this debate shows why I abhorr the SI units in theoretical electrodynamics. It confuses things in introducing unnecessary asymmetries into the equations of the field components, ##F_{\mu \nu}## (or equivalently ##(\vec{E},\vec{B})## in the (1+3) notation). The most natural units are rationalized Gaussian units (also known as Heaviside-Lorentz units). In these units ##\vec{E}## and ##\vec{B}## have the same dimension, and the fundamental Maxwell equations read
$$\vec{\nabla} \times \vec{E} + \frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0,$$
$$\vec{\nabla} \times \vec{B} - \frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{E}=\rho.$$
These are the microscopic Maxwell equations with ##j^{\mu}=(c \rho,\vec{j})## the complete charge-current density.
To derive electromagnetic waves for the free fields, i.e., ##\rho=0## and ##\vec{j}=0##, take the curl of the first equation
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{E})+\frac{1}{c} \partial_t \vec{\nabla} \times \vec{B}=0.$$
Now use the 3rd equation with ##\vec{j}=0##, and you get
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{E})+\frac{1}{c^2} \partial_t^2 \vec{E}=0.$$
Further you can work in Cartesian coordinates and then simplify the first term to
$$\vec{\nabla} (\vec{\nabla} \cdot \vec{E})-\Delta \vec{E} + \frac{1}{c^2} \partial_t^2 \vec{E}=0.$$
Using the fourth Maxwell equation with ##\rho=0## finally gives the wave equation
$$\left (\frac{1}{c^2} \partial_t^2-\Delta \right) \vec{E}=0.$$
This shows that there are solutions for ##\vec{E}## describing plane waves,
$$\vec{E}(t,\vec{x})=\vec{E}_0 \cos(\omega t-\vec{k} \cdot \vec{x}),$$
where
$$\omega=c|\vec{k}|,$$
i.e., a wave with phase velocity ##c##, i.e., the parameter ##c## in the Maxwell equation is the phase velocity of electromagnetic waves. The value of ##c## was known at Maxwell's times from electrostatic and magnetostatic (static!) measurements, and thus the Maxwell equations lead to a prediction (a) of the existence of electromagnetic waves and (b) a strong hint that light is an electromagnetic wave, because ##c## was pretty much the value measured by optical means.
As I said, the introduction of a fourth unit for charge (or as done officially in the SI electrich current) obscures this very clear picture a bit. In fact in the modern way to build the SI units it becomes very clear that ##\epsilon_0## and ##\mu_0## are just conversion factors from the arbitrary SI units to the more natural units (in the extreme you work in Planck units, where everything is expressed in dimensionless numbers ;-)): For the mechanical+electromagnetic part of physics the SI units are based on relativity, and there is one fundamental parameter, which is the speed of light, ##c##. Consequently ##c## is no longer measured but fixed exactly to define the unit of length (metre) in terms of the unit of time (second). Indeed, given relativity it's unnatural to measure space and time with different units, and that's why ##c## is just a conversion factor between metres and light seconds. This fixes the kinematical units of space and time. As a third unit for mechanics you need also a unit for mass, which is the kg in the SI.
In principle there's no need for a third unit of electric charge or current, and that's how the Gaussian or Heaviside-Lorentz units are built. Nevertheless, it leads to a bit of strange units for charges and currents with non-integeger powers of the base units for the electric charge and current. That's why in the SI they introduce the Ampere as a fourth unit. This requires a conversion factor, the "permeability of the vacuum", ##\mu_0=4 \pi 10^{-7} \text{N}/\text{A}^2##, where
##1 \text{N}=1 \text{kg} \text{m}/\text{s}^2## is the unit of force (Newton), and ##1 \text{A}## is the unit of the electromagnetic current, formally defined through the force per unit length of two infinitely long very thin wires in parallel at a distance of ##1 \text{m}##. Consequently (and since idiosyncrazily the units of the electric and the magnetic field components are also measured in different units!) you also need another conversion factor, the permittivity of the vacuum, defined by the Maxwell equation
$$\vec{\nabla} \cdot \vec{E}=\frac{1}{\epsilon_0} \rho.$$
In SI units the Maxwell equations lead to
$$c^2=\frac{1}{\mu_0 \epsilon_0},$$
which together with the definition of the speed of light through fixing the metre by the definition of the second and ##\mu_0## through the definition of the Ampere leads to the definition of ##\epsilon_0##. So indeed ##\mu_0## and ##\epsilon_0## are merely conversion factors from the artificial units of the SI to the more natural Gaussian or (even better) rationalized Gaussian units used earlier.
