I understand that one can measure a single photon being absorbed using a photomultiplier tube or CCD.
Can one measure a single photon being emitted by monitoring the current through an LED or the recoil of an emitting ion?
The photon is a particle. It will have particle interactions, i.e. scattering off electrons and/or the spill over electric and magnetic fields of solids, ( for example Compton scattering) , but the probability of ionization is very low, in contrast to a charged particle. Once it scatters inelastically its frequency and direction will change and no setup can be devised to be called "monitoring" because no current can be built up by one photon without drastically changing what one is trying to detect.
In the above bubble chamber photograph we see the following :
A gamma ray ( high energy photon) coming from the top and invisible (because of the low interaction rate), we only know it is there after it interacts by a scatter with an electron of an atom of the bubble chamber liquid.
In the first vertex we see an electron positron pair and and the recoil electron on which the gamma scattered. The second vertex is another electron positron pair, more energetic, by the scattered gamma, this time in the field of a nucleus, (because there is no third particle visible, the nucleus is much heavier than an electron and a lot of the gamma energy has been lost by the second scatter it has recoiled less than a few microns distance ).
Is it possible to detect the same photon both being emitted and later being absorbed?
As seen in the photo above, the process of detection involves interactions and the interactions change the setup. In the case of the energetic photon above we can say that it is the same photon creating both pairs because the setup for bubble chambers is constructed such that few gammas arrive for each picture , taken in coincidence with the production of the beam. The energies are such that a macroscopic "mapping" can be made, but one cannot identify the interaction that produced the gamma in the beam with the end gamma in the chamber. Only statistically.
The problem would be similar for low energy, and worse than seeing the two scatters here. Absorption means that an atomic level interaction has taken place and the photon kicked an electron up to a higher energy level. This is what one sees at the CCds and photos, absorption of the photons, but to connect with the source atom would be very hard, and not worth the effort since it is easy to do it statistically.
I suppose one can confirm that it is the same photon if the spacetime interval between the two events is zero.
The velocity of light is a well measured number , but not with single photons. It is a statistical measure.
Edit: the question expanded while I was replying
After the photon's emission has been detected, but before its absorption has been registered, is it real or virtual?
See the explanation of virtual in this answer. The photon is real as much as the electrons are real in the above photo. It just has a much lower probability of ionizing interactions and its track is not made visible the way the electrons' tracks are.
I assume a QED calculation would proceed on the assumption that the photon is virtual but if its emission has been measured then surely it should be treated as real?
A QED calculation pertains to the quantum mechanical framework, of nanometers and less . A Feynman diagram for the above picture would be written having as external lines the incoming gamma, a virtual gamma from the electron interacting with it and generating a real e+e- pair. The second interaction would have a nucleus emitting a virtual gamma which turns into an e+e- pair.
Z is the nucleus in the above simplest diagram.( There has to be a third particle involved because of momentum conservation). The gamma from Z is virtual, as is the intermediate electron. The e+e- are real. To describe the interaction in the bubble chamber photo, i.e. a gamma Compton scattering and creating an e+e- pair more virtual exchanges will be needed, but would not change the reality of the incoming and outgoing final gamma
Is there a paradox here?
No, because writing a single Feynman diagram for a macroscopic system has little meaning. What is virtual and what is real is a matter of framework and to describe macroscopic systems quantum mechanically one enters great complexity and has to use the density matrix formalism , not the simple interactions depictable by simple Feynman diagrams and the virtual exchanges there.