Can Maxwell's equations be derived from Coulomb's Law and Special Relativity?

Maxwell's equations do follow from the laws of electricity combined with the principles of special relativity. But this fact does not imply that the magnetic field at a given point is less real than the electric field. Quite on the contrary, relativity implies that these two fields have to be equally real.

When the principles of special relativity are imposed, the electric field $\vec{E}$ has to be incorporated into an object that transforms in a well-defined way under the Lorentz transformations - i.e. when the velocity of the observer is changed. Because there exists no "scalar electric force", and for other technical reasons I don't want to explain, $\vec{E}$ can't be a part of a 4-vector in the spacetime, $V_{\mu}$.

Instead, it must be the components $F_{0i}$ of an antisymmetric tensor with two indices, $$F_{\mu\nu}=-F_{\nu\mu}$$ Such objects, generally known as tensors, know how to behave under the Lorentz transformations - when the space and time are rotated into each other as relativity makes mandatory.

The indices $\mu,\nu$ take values $0,1,2,3$ i.e. $t,x,y,z$. Because of the antisymmetry above, there are 6 inequivalent components of the tensor - the values of $\mu\nu$ can be $$01,02,03;23,31,12.$$ The first three combinations correspond to the three components of the electric field $\vec{E}$ while the last three combinations carry the information about the magnetic field $\vec{B}$.

When I was 10, I also thought that the magnetic field could have been just some artifact of the electric field but it can't be so. Instead, the electric and magnetic fields at each point are completely independent of each other. Nevertheless, the Lorentz symmetry can transform them into each other and both of them are needed for their friend to be able to transform into something in a different inertial system, so that the symmetry under the change of the inertial system isn't lost.

If you only start with the $E_z$ electric field, the component $F_{03}$ is nonzero. However, when you boost the system in the $x$-direction, you mix the time coordinate $0$ with the spatial $x$-coordinate $1$. Consequently, a part of the $F_{03}$ field is transformed into the component $F_{13}$ which is interpreted as the magnetic field $B_y$, up to a sign.

Alternatively, one may describe the electricity by the electric potential $\phi$. However, the energy density from the charge density $\rho=j_0$ has to be a tensor with two time-like indices, $T_{00}$, so $\phi$ itself must carry a time-like index, too. It must be that $\phi=A_0$ for some 4-vector $A$. This whole 4-vector must exist by relativity, including the spatial components $\vec{A}$, and a new field $\vec{B}$ may be calculated as the curl of $\vec{A}$ while $\vec{E}=-\nabla\phi-\partial \vec{A}/\partial t$.

You apparently wanted to prove the absence of the magnetic monopoles by proving the absence of the magnetic field itself. Well, apologies for having interrupted your research plan: it can't work. Magnets are damn real. And if you're interested, the existence of magnetic monopoles is inevitable in any consistent theory of quantum gravity. In particular, two poles of a dumbbell-shaped magnet may collapse into a pair of black holes which will inevitably possess the (opposite) magnetic monopole charges. The lightest possible (Planck mass) black holes with magnetic monopole charges will be "proofs of concept" heavy elementary particles with magnetic charges - however, lighter particles with the same charges may sometimes exist, too.

Lubos Motl's answer is very good, but I think it's worth saying one or two additional things.

You can regard magnetism as simply a byproduct of electricity, in the following sense: if you assume that Coulomb's Law is correct, and that special relativity is correct, and that charge is a Lorentz scalar (so that charge and current density form a 4-vector), then you can derive all of Maxwell's equations. (Actually, you probably also need to assume the theory is linear as well, now that I think about it.) The undergraduate-level textbook by Purcell works this out very explicitly in a nice, pleasing way, and it's also in more advanced textbooks.

Some books gloss over the need to postulate that charge is a scalar. At least one textbook -- I don't remember which -- does emphasize it, and makes a convincing case that it's worth paying attention to. One way to see that it's not a trivial condition to impose is to consider the analogy with gravity -- that is, substitute mass for charge and gravity for electric field, and try to run the same argument. (Assume weak fields so that everything can be treated as linear if you like.) There are "gravitomagnetic" effects, but they're not related to regular gravity in the same way as the magnetic field is related to the electric field -- i.e., the gravitational analogues of Maxwell's equations look different from the regular Maxwell equations). One reason is the sign differences, of course -- like charges repel in one case and attract in the other. But a bigger reason is that the source of gravity is not a scalar: its density doesn't form part of a 4-vector, but rather of a rank-2 tensor.

But on a more philosophical (or perhaps semantic) level, I wouldn't jump from this fact to the conclusion that magnetism is "merely" a byproduct of electricity. At the very least, such language doesn't appear to be useful in understanding the theory or in using it! For instance, understanding how an electromagnetic wave can propagate from a distant galaxy to your eye is much easier and more natural if you look at it from the "usual" point of view.

Not a direct answer to your question but still a surprising derivation of Maxwell's equations:

Feynman's proof of the Maxwell equations (FJ Dyson - Phys. Rev. A, 1989) shows, that it is possible to derive Maxwell's equations from Newton's second law of motion and commutation relations (under non-relativistic limits).