Another option, besides modifying the potential $A_\mu = (A_i, \phi)$ in some way, is to introduce another 4-potential $C_\mu = (C_i, \psi)$.
Then the electric and magnetic field are given by
$$E = - \nabla \times C - \frac{\partial A}{\partial t} - \nabla \phi$$
$$B = \nabla \times A - \frac{\partial C}{\partial t} - \nabla \psi$$
More on this 2-potential approach can be found here: http://arxiv.org/abs/math-ph/0203043
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.
Best Answer
Yes it can be used,
Given,$ \nabla × A = B$
Then
$\nabla ×E = -\frac{\partial B}{\partial t}$
$\nabla × E = -\frac{\partial (\nabla × A) }{\partial t}$
$\nabla × E = -\nabla × \frac{\partial A }{\partial t}$
$\nabla × E + \nabla × \frac{\partial A }{\partial t} = 0$
$\nabla × (E + \frac{\partial A }{\partial t})= 0$
Because the curl of this quantity is zero, it can be written as the gradient of a scalar function ( or the negative of a gradient of a scalar function, which is used to match the definition of electrostatic potential)
$E + \frac{\partial A }{\partial t} = -\nabla V$
$E = -\nabla V -\frac{\partial A }{\partial t} $
From here we can substitute the definitions of E into the other maxwell equations
This will obtain 2 equations that interlink A and V in terms of the source terms $\rho$ and J.
using the "lorenz" gauge choice ( not lorentz) we can decouple these and solve easier( or using the coulomb gauge)
This is called the potential formulation of maxwells equations