Does a single electron moving at velocity $v$ have an associated magnetic field, ignoring intrinsic spin? I have seen explanations of the magnetic field due to an electric current as being due to a Lorentz contraction of the moving electric charges. Would this explanation work for a single electron. There is still a current associated with a single moving electron. 
 A: 
I have seen explanations of the magnetic field due to an electric
  current as being due to a Lorentz contraction of the moving electric
  charges.

I'm afraid that's not quite the correct understanding.  The magnetic field, due to an electric current, is due to the moving electric charges, period.
What we need to do here is be careful to distinguish between magnetic field and magnetic force.
(1) If a charged particle is at rest in a non-zero magnetic field, there will be no magnetic force on the particle.
To emphasize, only if the charged particle is moving in a non-zero magnetic field will there be a magnetic force acting on the particle.
I think where you've picked up the idea in the above quote is in the explanation for the force on a charged test particle moving with respect to a current carrying wire.
In the rest frame of the wire, the wire is uncharged and thus, the force on the charged particle is entirely magnetic (due to the magnetic field of the current).
However, in the rest frame of the test particle, by (1) there can be no magnetic force acting.
Thus the force on the particle, in this frame, must be entirely electric and this implies that, in this frame, the current carrying wire has a net electric charge density.
How does this charge density appear in this frame?  The fixed charge density and moving charge density (the current) in the wire are changed, due to Lorentz contraction, in this frame from their values in the rest frame of the wire.

So, your question is based on, I think, a misunderstanding.  A moving electric charge has an associated magnetic field since electric current is charge in motion.
To think of this in terms of relativity, simply recognize that the electric scalar potential and magnetic vector potential are components of a four-vector, the four-potential, and thus the components mix together under a Lorentz transformation.
So, a four-potential that is purely electric in character in one frame, will, in general, be both electric and magnetic in nature in other, relatively moving frames.
A: The electromagnetic field created by a moving relativistic charged point particle is given by the Liénard-Wiechert potential. They are slightly more complicated than this answer really gives room for but they are essentially the solutions for the Maxwell equations in terms of potentials,
$$\partial_\mu\partial^\mu A_\nu=-\mu_0 J_\nu,$$
where you take the current density four-vector corresponding to a single particle. Thus if in a given inertial frame the particle's trajectory is characterized by $r(\tau)=(t(\tau),x_j(\tau))$, the four-current density is given by 
$$J_\nu(x)=ec\int\mathrm{d}\tau \frac{\mathrm d r_\nu}{\mathrm d\tau}\delta^{(4)}(x-r(\tau))
$$
(cf. Jackson, section 14.1). One standard approach to this is to solve the problem via Green functions, and therefore first solve the fundamental equation
$$\partial_\mu\partial^\mu D(x,x')=\delta^{(4)}(x-x').$$
(This is physically very weird: it represents the scalar field of a charged particle that only exists at one point in space and at one instant in time. The solution is essentially a Coulomb field in the fictional particle's rest frame, propagated along the particle's forward (or backward?) light cone; see Jackson, section 12.11.) You then convolute this with the source to get
$$A_\nu(x)=-\mu_0\int\mathrm d^4 x'D(x,x')J_\nu(x')=-\mu_0ec\int\mathrm d\tau D(x,x')  \frac{\mathrm d r_\nu}{\mathrm d\tau}.$$
This solution winds up depending heavily on the concept of retarded time: given a point $x$, the retarded time $\tau_\mathrm{ret}(x)$ is the (unique) time $\tau$ such that $x$ is in the forward light cone of $r(\tau_\mathrm{ret}(x))$ - and thus the point in the trajectory $r(\tau)$ which will radiate to $x$. While this is fine, it gets very messy when you want to get the electric and magnetic fields, as you need to differentiate $\frac\partial{\partial x^\mu}\tau_\mathrm{ret}(x)$, and that is a pretty painful experience.
If you have specific questions, or this seems much too complicated, I'm happy to elaborate or simplify.
