Can QED be formulated in terms of independent vector potential and bivector fields? We can obtain the classical equations of motion for electromagnetism by considering the vector potential $A^\mu$ and the electric $\mathbf E$ and magnetic $\mathbf B$ fields as independent degrees of freedom.
$$\mathcal L = j^\mu A_\mu - A^0 (\nabla \cdot \mathbf E) - \mathbf A \cdot (\dot {\mathbf E} - \nabla \times \mathbf B) - \frac12 (\mathbf E^2 - \mathbf B^2).$$
Varying the action with respect to $A^\mu$ yields the inhomogeneous Maxwell equations in terms of $\mathbf E$ and $\mathbf B$ (up to possible sign errors):
$$-\frac{\delta S}{\delta A^0} = \nabla \cdot \mathbf E - j^0$$
$$-\frac{\delta S}{\delta \mathbf A} = \dot {\mathbf E} - \nabla \times \mathbf B + \mathbf j$$
while varying with respect to $\mathbf E$ and $\mathbf B$ yields the standard definitions of those fields in terms of derivatives of $A^\mu$:
$$-\frac{\delta S}{\delta \mathbf E} + \mathbf E = \dot {\mathbf A} - \nabla A^0$$
$$\frac{\delta S}{\delta \mathbf B} + \mathbf B = \nabla \times \mathbf A.$$
Under a gauge transformation $A_\mu \to A_\mu + \partial_\mu \lambda$, $\mathcal L$ is not invariant, but $\delta S/\delta \lambda \equiv 0$ when $\partial_{\mu}j^{\mu} = 0$.
In this picture, $A^\mu$ acts as an intermediary between the bivector field $(\mathbf E, \mathbf B)$ and the source of the current density $j^\mu$ (e.g. a spinor field).  The relations between $(\mathbf E,\mathbf B)$ and the exterior derivative of the vector potential arise as classical equations of motion, rather than by definition.


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*Are there problems with this formulation that invalidate it at the classical level?  (e.g. too many degrees of freedom, problems with a Hamiltonian formulation, etc.)  It looks like the Hamiltonian wouldn't be bounded from below, but maybe there's a workaround.

*Does this setup have an analogue in quantum field theory, where we normally consider the gauge field $A^\mu$ as the only fundamental degrees of freedom for electromagnetism?
 A: I) OP's action in covariant notation$^1$
$$ S_1[A,F] ~:=~\int \! d^4x~{\cal L}_1, \qquad {\cal L}_1~:=~\frac{1}{4}F^{\mu\nu}F_{\mu\nu}- F^{\mu\nu}\partial_{\mu} A_{\nu} + j^{\mu}A_{\mu}, $$ 
$$ E_i~\equiv~F_{i0}, \qquad B_i~\equiv~\frac{1}{2}\epsilon_{ijk}F_{jk}, \tag{A} $$
is the first-order/Palatini formulation of E&M, cf. Ref. 1 & comment by Cosmas Zachos.
It is classically well-defined. The EL eqs. for the OP's action (A) read
$$ F_{\mu\nu}~\approx~\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu},  \tag{B}$$
$$ d_{\mu} F^{\mu\nu}+j^{\nu}~\approx~0.\tag{C}$$
The quadratic potential term 
$${\cal V}~=~-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+\ldots~=~\frac{1}{2}({\bf E}^2-{\bf B}^2)+\ldots\tag{D}$$ 
has a minus sign in front the independent ${\bf B}$-field, and is hence unbounded from below. Therefore OP's action (A) is quantum mechanically ill-defined. 
II) However, if we integrate out the independent ${\bf B}$-field, we basically get the Hamiltonian formulation of E&M, 
$$ S_H[A,{\bf E}] ~:=~\int \! d^4x~{\cal L}_H, \qquad {\cal L}_H~:=~ -{\bf E}\cdot \dot{\bf A}-{\cal H}, $$ $$ {\cal H}~:=~\frac{1}{2}({\bf E}^2+(\nabla \times {\bf A})^2)-{\bf J}\cdot {\bf A} +A_0{\cal G} \qquad {\cal G}~:=~\nabla \cdot {\bf E}-\rho, \tag{E}$$
cf. Ref. 2, which is quantum mechanically well-defined. (Minus) the independent ${\bf E}$-field plays the role of momentum for the magnetic gauge potential ${\bf A}$. Moreover, $A_0$ becomes the Lagrange multiplier for Gauss' law 
$${\cal G}~\approx~0.\tag{F}$$ 
To achieve QED, one should then proceed with quantization, gauge-fixing, etc.
References:


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*ADM, arXiv:gr-qc/0405109; eq. (3.5).

*P.A.M. Dirac, Lectures on Quantum Mechanics, 1964; chapter 2.
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$^1$ We use signature convention $(−,+,+,+)$ and $c=1$. Disclaimer: In this answer we have ignored some total space-time divergence terms in the action as they don't contribute to EL eqs.
