In 3-space, one can interpret the 4 Maxwell equation as determining the relationship between the fields (the electric field vector and the magnetic field bivector) and all four types of possible sources.
But this is rather illusory. In relativity, the equations look quite different:
$$\begin{align*} \nabla \cdot F &= -\mu_0 J \\ \nabla \wedge F &= 0\end{align*}$$
where $F$ is the electromagnetic field bivector. The vector derivative $\nabla$ can only increase or decrease the grade of an object by 1. Since $F$ is grade 2, the divergence equation describes its relationship with a grade 1 source term (the vector four-current $J$). The curl equation describes its relationship to a grade 3 (trivector) source term (of which there is none).
The reason the 4 Maxwell equations in 3-space come out the way we do is that we ignore the timelike basis vector, which would unify the scalar charge density with the 3-current as the four-current, as well as unify the E field with the B field as a bivector. The relativistic formulation, however, is considerably more sensible, as it correctly presents the EM field as one object of a single grade (a bivector), which can only have two sources (vector or trivector). It just so happens that the EM field has no trivector source.
What if there were trivector sources? Well, as you observe, there would be magnetic charge density (monopoles), but there would also be quite a bit more. There would have to be magnetic current as well, which would add an extra term to the $\nabla \times E$ equation to fully symmetrize things.
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.
Best Answer
Your problem is that you did not take relativity into account:
In Minkowski space, the relation between exterior derivatives and classical vector operators is different from the one in Euclidean 3-space, and $E$ and $B$ actually turn out to be components of a single 2-form $F$ (which is necessary to get the correct transformation laws under boosts).
Because I'm lazy, I'm going to work backwards from ${\rm d}F$ and ${\rm d}\star F$.
First, the electromagnetic tensor can be decomposed into $$ F = \sum_i E_i\,{\rm d}t\wedge{\rm d}x^i - \star\sum_i B_i\,{\rm d}t\wedge{\rm d}x^i $$ I'm assuming a $(+---)$ convention for the Minkowski metric. Please note that the sign above might be incorrect - I know I messed up somewhere (I started out with a $+$ in the formula above, and 'fixed' it after I got the wrong result), so it might be a good idea for someone to check these calculations and correct my answer if they are wrong.
The exterior derivative on 2-forms can be written as $$ \begin{align*} {\rm d}\sum_i A_i\,{\rm d}t\wedge{\rm d}x^i &= \star\sum_i (\nabla\times A)_i\,{\rm d}x^i \\ {\rm d}\star\sum_i A_i\,{\rm d}t\wedge{\rm d}x^i &= -\star(\nabla\cdot A\,{\rm d}t + \sum_i \frac{\partial A_i}{\partial t}\,{\rm d}x^i) \end{align*} $$ and we arrive at $$ \begin{align*} {\rm d}F &= \star\sum_i (\nabla\times E)_i\,{\rm d}x^i + \star(\nabla\cdot B\,{\rm d}t + \sum_i \frac{\partial B_i}{\partial t}\,{\rm d}x^i) \\&= \star\sum_i ( \nabla\times E + \frac{\partial B}{\partial t} )_i\,{\rm d}x^i + \star\nabla\cdot B\,{\rm d}t \\ {\rm d}\star F &= {\rm d}\left( \star\sum_i E_i\,{\rm d}t\wedge{\rm d}x^i + \sum_i B_i\,{\rm d}t\wedge{\rm d}x^i \right) \\&= -\star(\nabla\cdot E\,{\rm d}t + \sum_i \frac{\partial E_i}{\partial t}\,{\rm d}x^i) + \star\sum_i (\nabla\times B)_i\,{\rm d}x^i \\&= \star\sum_i ( \nabla\times B - \frac{\partial E}{\partial t} )_i\,{\rm d}x^i - \star\nabla\cdot E\,{\rm d}t \end{align*} $$ from which we get the left-hand sides of the Maxwell equations by looking and space and time components separately.