Thus we can see that the Al is oxidised but there are no Cu2+ ions to
be reduced also Na+ reduction would lead to a negative Ecell standard
and a positive Gibbs. Is it an air cathode where O2 is being reduced?
Well, there is no free lunch in this world. This is not a conventional cell in which each electrode is sitting in its own ions and one can easily calculate their Nernst potential. What we are probably seeing here is the very concept of contact potential. When two dissimilar metals are brought into contact via some electrolyte, a potential difference appears between the two. Recall the first experiment of Volta. He made a long series of
-silver[wet cardboard]zinc-silver[wet cardboard]zinc-silver[wet cardboard]zinc- cells, where the cardboard was wetted with salt-water. He was able to generate large potential differences this way by making very long series of Ag-Zn couple. Is this set-up good for drawing current-probably not.
These experiments were done in the 17th century by Volta. Atoms and ions did not exist then. These are contact potentials and you do not need a redox system to begin the experiment (metal dipping in its salt). Let us say you start with Al, Al will start to dissolve in a corrosive medium (salt is very corrosive) to form Al ions. You have a valid point, there are no copper ions to reduce? What else do we have in the system? Water, it has some ionized $\ce{H+}$, they start to reduce, one should see hydrogen bubbles at the copper electrode. Certainly you can draw small currents to power a LED from such batteries but not for long.
I do have some doubt about the design of the experiment. All soda cans are lined with a very very thin polymer layer. So the contact of Al is not that good with the copper coil. If one scrubs the can very well, only then that polymer layer would come off. The claim of seven LEDs is a stretch.