The idea that current can't flow at all in a circuit with an "open" (such as a capacitor or a switch) is an oversimplification given to introductory students. A more accurate statement is that constant, sustained current can't flow in a circuit with an open.
(The following description is classical. I'm sure an electrical engineer or quantum physicist would cringe at this explanation. Still, I think it's close enough to answer your question.)
In this edited picture from your question, the red dot represents the voltage at the output of the generator. Since this is AC, the voltage at the red dot will be varying continuously. In classical terms, that point will have a constantly changing surplus or deficiency of electrons.
Now, in an idealized circuit, the voltage at the blue dot would have the same voltage because "it's the same point electrically."
But no change actually happens instantaneously. In a real circuit, it will take some time for the electrons to migrate along the wire. The generator will be "pushing electrons" onto the area of the red dot, making it more negative than the charge at the blue dot. That difference in charge causes negatively charged electrons to repel each other, causing some amount of current to flow until enough electrons have reached the blue dot, so that the charge distribution is even along the length of wire between the red and blue dot. Once charge is distributed evenly along the length of wire, flow of current will cease.
But that is the trick. Because this is AC, charge will never be evenly distributed along the length of the wire. The charge at the blue dot will forever by lagging behind the amount of charge at the red dot because it takes an extremely small, but finite time for the charge to equalize between those two points. This is the idea behind the "time constant" you may have been introduced to in your studies of capacitance.
And that's the idea of the capacitor: The large plates that are rolled up inside the capacitor are meant to act as a reservoir of electrons. It takes some time to "charge" plates with excess electrons. And after the plates are fully charged, it takes some time for the extra electrons to drain off of them when the voltage reverses. That's why capacitors are said to oppopse a change in voltage: When voltage reverses—for a very short time—the electrons coming off one of the capacitor plates can make up for electrons the generator is no longer providing.
You would see the same thing (almost) in a DC circuit: Once a voltage source is applied to one end of the capacitor, it will take some time (roughly 5 "time constants") for enough electrons to move from the voltage source to the capacitor plate for change distribution to equalize. At that point, current will stop flowing because the capacitor is acting like an open.
So don't think that capacitors allow current flow in the same way that complete, continuous circuits allow current flow. Current never flows "though" a capacitor in the same way it flows through a closed circuit. A capacitor truly is an "open." But on extremely small time scales, some current will flow nonetheless because electrons repel each other, so they will flow along the wire, either towards or away from the capacitor, in order to equally distribute charge. And when AC is used (provided the frequency is fast compared to the capacitance), the capacitor will always be in the process of charging or discharging, so it looks like current flows through the capacitor.
With regard to the circuit diagrams in the original post: Will the light bulb light? To be pedantic, the only way to know would be to know the capacitance of the capacitor, the capacitance of the switch, the resistance of the bulb, and the frequency of the AC. Then run through the impedece calculations. To be practical (if limiting the discussion to residential wiring) we know from experience that opening a switch does in fact turn off a light. So we know that the capacitance of the switch is so vanishingly small that the displacement current isn't sufficient to power the light. And that makes sense. The plates of capacitors are designed to amass electric charge. The poles of switches are designed to minimize capacitance.
A word on jargon: It's very common for non-engineer technicians like me to be taught that current actually flows through a capacitor. (They are often taught by other non-engineers who are just passing on what they were taught.) I think that is more oversimplification since, for all intents and purposes, that how it acts for a technician who doesn't have to actually design circuits. Even engineers will talk about the capacitor "passing" high frequencies, but I think that is either speaking metaphorically or just engineer jargon. So you will find textbooks and other authorities speaking of AC current flow "through" a capacitor. The jargon may be so ingrained that they just don't think about it any more, so they may not be aware of the confusion that their wording can cause.
Pulling out my old Navy Electricity and Electronics Training Series Module 2 on alternating current, it has this to say (printed page number 4-8):
Since the plates of the capacitor are changing polarity at the same
rate as the ac voltage, the capacitor seems to pass an alternating
current. Actually, the electrons do not pass through the dielectric,
but their rushing back and forth from plate to plate causes a current
flow in the circuit. It is convenient, however, to say that the
alternating current flows "through" the capacitor. You know this is
not true, but the expression avoids a lot of trouble when speaking of
current flow in a circuit containing a capacitor. (p. 124 of 250 of the online PDF)