No explicit complexification is needed to derive this breakdown of Maxwell's equations. This can be understood wholly through the real vector space of special relativity.
Let's start with Maxwell's equations for the EM field, in the clifford algebra language called STA: the spacetime algebra. Maxwell's equations take the form
$$\nabla F = -J$$
where $\nabla F = \nabla \cdot F + \nabla \wedge F$, $F = e_0 E + B \epsilon_3$, in the $(-, +, +, +)$ sign convention.
Let $x$ be the spacetime position vector. It's generally true that, for a vector $v$ and a constant bivector $C$,
$$\nabla (C \cdot x) = -2 C, \quad \nabla (C \wedge x) = 2C \implies \nabla (Cx) = 0$$
One can then evaluate the expression
$$\nabla (Fx) = (\nabla F)x + \dot \nabla (F \dot x)$$
where the overdot means that only $x$ is differentiated in the second term; using the product rule, $F$ is "held constant" and so the above formulas apply. We just argued that the second term is zero, so we get $\nabla (Fx) = (\nabla F) x$. Thus, we arrive at the following transformation of Maxwell's equations:
$$\nabla (Fx) = -Jx$$
Now, we could always write $F$ as a "complex bivector" in the sense that, using $\epsilon = e_0 \epsilon_3$, and $\epsilon \epsilon = -1$, we have
$$F = e_0 E - B \epsilon_3 e_0 \epsilon_3 \epsilon = e_0 (E + \epsilon B)$$
It's crucial to note that $\epsilon$ does not commute with any vector.
What are the components of $Fx$? Write $x = t e_0 + r$ and we can write them as
$$Fx = e_0 (Ex + \epsilon Bx) = e_0 (E \cdot r + E \wedge r - e_0 Et + \epsilon B \cdot r - e_0 B \times r + \epsilon B t e_0)$$
This too can be written in a "complex" form:
$$Fx = (e_0 E \cdot r + Et + B \times r) + \epsilon (E \times r + e_0 B\cdot r + Bt)$$
We seem to differ on some signs, but this is recognizably the same quantity you have called $G$.
Now, to talk about how these equations break down, let's write $G = G_1 + G_3$, where $G_1 = (e_0 E \cdot r + \ldots)$ and $G_3 = \epsilon (E \times r + \ldots)$. Let's also write for $R = Jx = R_0 + R_2$.
Maxwell's equations then become
$$\nabla \cdot G_1= R_0, \quad \nabla \wedge G_1 + \nabla \cdot G_3 = R_2, \quad \nabla \wedge G_3 = 0$$
The first and third equations are the components of the Gauss dipole; the second equation is the Ampere-Faraday dipole equation.
Now, what does it all mean? The expression for $G = Fx$ includes both rotational moments of the EM field as well as some dot products, so it measures both how much the spacetime position is in the same plane as the EM field as well as how much the spacetime position is out of the plane.
It's probably more instructive to look at the source term $-Jx$. This tells us both about the moments of the four-current as well as how it goes toward or away from the coordinate origin. The description for the moments is wholly in the Ampere-Faraday dipole equation. What kinds of moments would this describe? A pair of two opposite point charges at rest, separated by a spatial vector $2 \hat v$ and centered on the origin, each with current at rest $j_0$, would create a $R = Jx = + j_0 e_t \hat v - j_0 e_t (-\hat v) = 2 j_0 e_t \hat v$, so this would be described wholly by the A-F dipole equation.
That's at time zero, however. At later times, $R$ will pick up these weird time terms. Say we're at time $\tau$. Then $R = 2 j_0 e_t \hat v + j_0 e_t (\tau e_t) - j_0 e_t (\tau e_t)$. So for this case, there's no problem: the extra stuff will just cancel. A single charge, however, would start picking up this term.
In a few words, these equations are weird.
In electrostatics, when you write the multipole expansion of the potential you find
\begin{equation}
\Phi(\vec{x})=\frac{1}{4 \pi \epsilon_0} \left[ \frac{q}{r} + \frac{\vec{p} \cdot \vec{x}}{r^3} + ... \right] \, ,
\end{equation}
where $r=|\vec{x}|$ and $...$ indicates the higher order multipole terms. We can find the effective charge distribution with
\begin{equation}
\nabla^2 \Phi = - \frac{\rho}{\epsilon_0} \, .
\end{equation}
Now we just need appropriate expressions for the laplacian of the terms we have. Remember that
\begin{equation}
\frac{\vec{x}}{r^3} = - \nabla \frac{1}{r} \quad \mathrm{and} \quad \nabla^2 \frac{1}{r} = - 4 \pi \, \delta(\vec{x}) \, .
\end{equation}
Now we will be able to reproduce the result if we notice that
\begin{eqnarray}
\nabla^2 \left[ \frac{\vec{p} \cdot \vec{x}}{r^3}\right] &=& \nabla \cdot \left[ \nabla \left( \frac{\vec{p} \cdot \vec{x}}{r^3}\right) \right] = \nabla \cdot \left[ (\vec{p} \cdot \nabla) \left( \frac{\vec{x}}{r^3}\right) \right] = \vec{p} \cdot \nabla \left[ \nabla \cdot \left(\frac{\vec{x}}{r^3}\right) \right]
\\
&=& - \vec{p} \cdot \nabla \left[ \nabla \cdot \left( \nabla \frac{1}{r}\right) \right]
= - \vec{p} \cdot \nabla \left[ \nabla^2 \frac{1}{r} \right] = 4 \pi \, \vec{p} \cdot \nabla \delta (\vec{x}) \, .
\end{eqnarray}
Check each equality carefully. Hence we find the effective charge density
\begin{equation}
\rho(\vec{x}) = q \, \delta(\vec{x}) - \vec{p} \cdot \nabla \delta(\vec{x}) \, + \, ... \, .
\end{equation}
EDIT:
Alternatively you can say
\begin{eqnarray}
-4\pi\nabla \delta(\vec{x}) &=&\nabla \left[ \nabla \cdot \left(\nabla \frac{1}{r}\right)\right]=\nabla \left[ \nabla \cdot \left(- \frac{\vec{x}}{r^3}\right)\right] = \nabla \times \left[ \nabla \times \left(- \frac{\vec{x}}{r^3}\right) \right] + \nabla^2 \left(- \frac{\vec{x}}{r^3}\right)
\\
&=& - \nabla^2 \left( \frac{\vec{x}}{r^3}\right) \, .
\end{eqnarray}
Now I think everything is consistent, sorry for the confusion.
Best Answer
Energy of a dipole $p$ is minus $p$ times electric field (https://unlcms.unl.edu/cas/physics/tsymbal/teaching/EM-913/section4-Electrostatics.pdf), so your symbol probably denotes the electric dipole moment (which is closely related to polarization, as @ZeroTheHero suggested).
EDIT (Jan 20, 2022) Looks like the derivation of the energy of an electric dipole in the electric field is given in Journal of Modern Optics (2004) vol. 51, no. 8, 1137–114, Section 2, so the symbol is indeed the electric dipole moment.