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.
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.
Best Answer
When $\frac{\partial E}{\partial t}=0$, then the 1st equation is valid. That is, it is for magnetostatics, where currents (and fields) are not time-varying.