When the electron strikes the anode there are many ways it can dissipate it's 24eV worth of energy. It could eject an electron from some atom in the anode, and that excited atom could emit a photon as it recombines with an electron, however it is vastly more likely that the energy will end up as lattice vibration i.e. heat. The amount of radiation emitted from the anode is very small, but it does happen and is used as the basis of soft x-ray emission spectroscopy.
In any case 24eV would be an exceedingly soft X-ray, if it could be called an X-ray at all. At that energy any radiation would be rapidly absorbed by pretty much everything including the glass of your valve.
Re your second question, in the absence of the anode the photoemission would charge the cathode and increase the work function. So you'd get some photemission as first, but it would stop after a while. With the anode I guess the electrons would indeed arrive with more than 24ev energy.
The point "I" marks the transition from a low-current glow discharge, where the gas characteristics determine the behavior, to a high current arc, where electrode thermionic or high-field emission and even evaporation dominate.
Below is a "first-order" analysis of the transitions at points "D" and "I".
At low levels of current ("B" - "D"), what's happening here is that an electron emitted from the cathode collides with and ionizes gas atoms on its way to the anode, producing additional electrons, which can ionize additional atoms, etc. For $n_0$ initial electrons emitted by the cathode, with the cathode-anode spacing $d$, the number of electrons reaching the anode $n_a$ is:
$$ n_a = n_0 e^{\alpha d} $$
Here $\alpha$ is the first Townsend ionization coefficient; $1/\alpha$ is the average distance between ionizing collisions. On average, each emitted electron produces $ \left( e^{\alpha d} - 1 \right)$ new electrons.
It turns out that:
$$ \frac{\alpha}{p} = f \left( \frac{E}{p} \right) $$
where $p$ is the gas pressure, $E$ is the applied electric field, and $f$ denotes a functional relationship. (So at constant pressure, as the applied voltage increases, $\alpha$ does also.)
Of course, each ionized gas atom results in a positively charged ion as well as an electron; these ions drift back to the cathode. (Total current $I=I_0e^{\alpha d}$ is the sum of the electron and ion currents across any plane between the electrodes.)
What happens when these ions reach the cathode is crucial to the dark current / glow discharge transition. When ions collide with the cathode, some fraction $\gamma$ will "kick out" a new electron from the cathode; $\gamma$ is the second Townsend ionization coefficient. With this effect included, the total number of electrons emitted from the cathode $n_{0t}$ is:
$$ n_{0t} = n_0 + \gamma n_{0t} \left(e^{\alpha d} -1 \right) $$
Note the feedback effect now present: each electron emitted by the cathode can produce additional cathode electrons via the two effects. Solving:
$$ n_{0t} = n_0 \frac{e^{\alpha d}}{ 1 - \gamma \left(e^{\alpha d} -1 \right)} $$
$\gamma$ is also a (different) function of $E/p$:
$$\gamma=g (E/p)$$
From this formula one can identify the Townsend criterion for a spark breakdown, where the total number of electrons (and the current) go to infinity:
$$ \gamma \left( e^{\alpha d} -1 \right) = 1 $$
What happens next depends on the source of electricity. If we were gradually increasing the voltage of a "stiff" source (one with low series resistance) connected directly across the tube, the "infinite" current when the Townsend criterion is met, would run away directly to the arc region beyond "I", bypassing the glow discharge region entirely.
Suppose instead that we have a fixed voltage source connected to the tube via a large variable series resistor. (See this nice analysis, for example.) The operating point is varied by adjusting the resistor. Then, after breakdown at "D", the tube can operate stably within the glow discharge region, with the tube operating point automatically adjusting to the source (voltage supply + variable resistor). [The operating point is stable if the total resistance (variable resistor + incremental resistance of tube) is positive.]
The glow discharge current can be increased by reducing the external variable resistor. However, glow discharge current cannot be increased indefinitely. As the tube current and voltage increase, the bombardment of that current heats the electrodes, leading to increased thermionic emission from the cathode. With sufficient heating (i.e. at point "I"), cathode thermionic emission and material evaporation creates a lower-voltage "alternative" to the glow discharge, and a full-blown arc results. In this mode, the original gas in the tube is in a sense redundant; the arc can create its own plasma from vaporized electrode material.
[Note: there appears to still be uncertainty about the relative importance of thermionic and field emission in arc behavior.]
Best Answer
Once the positive gas ion hits the cathode (= roughly an infinite bucket of electrons) it neutralizes and flows back into the tube as neutral gas. You're right that otherwise the discharge would be quite short-lived.