It is possible to generate a single flying electron (S. Hermelin, et.al., "Electrons surfing on a sound wave as a platform for quantum optics with flying electrons", Nature, 21 Sept 2011, 477 (7365)).
What is the mechanism by which such a single electron moving in free space generates a magnetic field?
It fundamentally is a relativistic effect. Do you accept that a stationary electron generates an electric field which can be measured? I'll assume it is okay and build upon that.
A stationary electron creates an electric field with $$\vec E(\vec x) \propto \frac{q}{|\vec x - \vec x_0|} (\vec x - \vec x_0) \,.$$ Let us choose $\vec x$ and $\vec x_0$ such that they only are along the $z$-axis. So say we are at the origin and the electron is at $\vec x_0 = (0, 0, 1)$. Then we would see some $E_z$ field component. Since the electron is at rest with respect to us, we don't see any magnetic field.
This electric field can also be written in terms of the field strength tensor. For this we use $E^i = F^{i0}$. So the tensor is roughly this $$ \mathsf F = \begin{pmatrix} 0 & 0 & 0 & E_z \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -E_z & 0 & 0 & 0 \end{pmatrix} \,.$$ There can be an overall sign depending on where one puts the spacetime indices. It does not matter here as this is just a qualitative argument.
We now make a change of coordinate systems. We move along the $x$ axis and keep the electron as it is. It will appear that the electron moves relative to us now. The spacetime transformation will look like this $$ \mathsf \Lambda = \begin{pmatrix} \cosh \rho & \sinh \rho & 0 & 0 \\ \sinh \rho & \cosh \rho & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \,.$$
Performing the transformation which is roughly $\mathsf F \to \mathsf \Lambda \mathsf F \mathsf \Lambda$, we obtain a new field strength tensor which now looks like this: $$ \mathsf F' = \begin{pmatrix} 0 & 0 & 0 & E_z \cosh \rho \\ 0 & 0 & 0 & E_z \sinh \rho \\ 0 & 0 & 0 & 0 \\ -E_z \cosh \rho & -E_z \sinh \rho & 0 & 0 \end{pmatrix} \,.$$
The new elements there correspond to the $B_y$ components. So there is now a magnetic field in the $y$ direction.
If you want to ask how the electric field is produced, it will probably not shed that much light by looking into QED. I would think that there is no really more fundamental explanation for the electric field.
At some level, physics can't explain anything. It can only describe how nature works. We can "explain" things if you are willing to accept some set of axioms that remain unexplained, and then reason logically from those axioms. But since the axioms themselves are unexplained ... have you really managed to explain anything?
If you can tell us what you are willing to accept, we might be able to give you an "explanation". For example, are you willing to accept that EM fields can be represented by vectors, and that Maxwell's equations are valid? If so, we can help.
There are several levels of explanation to most physics questions.
One level is: Any electric current generates a magnetic field. A moving electron is a (very small) electric current. If you take a zillion electrons moving along a wire (you don't really need the wire), then you have a magnetic field as you can confirm by experiment. Each electron contributes a zillionth of the total magnetic field. If you take away a zillion minus 1 electrons from the system, you are left with the zillionth-size magnetic field.
Another level: If you have an electric field in a certain frame of reference, then in another frame of reference you have a magnetic field (and vice versa). This is due to the way an electromagnetic field transforms in Relativity. The electron doesn't know that it is moving, and in its frame it just has a pure electric field without any magnetism. You, however, are in a different frame from the electron, and so you see a magnetic field.
Another level: It is perfectly valid to go into the electron's frame and calculate what happens when your experimental apparatus, moving at high speed past the electron, encounters the electron's pure electric field. The results of that calculation will agree with what you calculate when you treat the apparatus as stationary and interacting with the moving electron's electric + magnetic field. However, one way of doing the calculations is quite a bit easier than the other, and laziness is a virtue.
The answer depends on your starting point for what you consider to be unnecessary to explain. For example, if you think it is unnecessary to explain why a stationary electric charge produces a static electric field that acts like straight lines emanating from the electron, and if you think the rules of how 4-vectors transform when you change reference frames need no explanation, then you can say that the magnetic field is the natural consequence of taking a static electric field and observing it from a moving frame of reference. This allows the electric field to be placed into the larger context of the electromagnetic tensor (see https://en.wikipedia.org/wiki/Electromagnetic_tensor), whereby we can regard both electric and magnetic fields in a unified way in which they both stem from a 4-potential. That's similar to how relativity allows us to view space and time as part of a unified 4-dimensional spacetime. So if we answer your question this way, we say it needs to produce a magnetic field such that what you already understand about electric fields can be placed into the broader context of how the universe behaves when you change reference frames. That relativity is needed is evidenced by the fact that $c$ appears whenever you start talking about the $B$ field of a moving charge.
