Definition of ionisation energy
It is common practice to define the standard state of a gas when describing its thermodynamic properties. For instance, the free energy of an ideal gas can be defined relative to the free energy in its standard state (the gas at a partial pressure of 1 bar and at the same temperature), as 

$$G = G^\circ + RT\log\left( \frac{p}{p^\circ} \right)$$

where $p^\circ= \pu{1 bar}$.

In practice it may be impossible to achieve the gaseous state described in the OP, not least because absolute zero is not experimentally accessible. However, there are workarounds that might allow one to extrapolate to $\pu{0 K}$ and to standard conditions (a pressure of 1 bar). First the easy part, the pressure: if you have a very dilute gas, then intermolecular interactions become unimportant in describing its behavior, it becomes practically ideal. As such, its behavior at any other pressure can be described by using standard equations applicable to ideal gases. It does not matter that the state described is not real, it can be used as a hypothetical reference point. For instance, if we know the free energy under ideal gas conditions at a particular pressure, we can compute the free energy in the standard state as follows:   

$$G^\circ = G - RT\log\left( \frac{p}{p^\circ} \right)$$

For an ideal gas, the internal energy $U$ is independent of pressure, so in the given example there is no need to perform any computation at all, provided we are at the right T. The problem is of course that we are not at the right T. In practice it may be possible to either assume that the experiment was conducted sufficiently close to $T=\pu{0 K}$, so that we can assume the equality holds, or to extrapolate to that temperature by performing experiments over a range of T. 

As for the approximation $\Delta U (\pu{0 K})\approx \Delta H (\pu{298 K})$, I'll give a roundabout justification, since I am not an expert on the subject. There are two points that can be made. First, it is probably ok to assume $\Delta U (\pu{0 K})\approx \Delta U (\pu{298 K})$ if it can be assumed that the electronic transition is not sensitive to differences in the distribution of (mainly vibrational) thermal motions at $\pu{0 K}$ and $\pu{298 K}$. 

Second, $$\Delta H = \Delta U + \Delta (pV) = \Delta U + \Delta n RT $$ However here we have a conundrum. The reported IE ignores the kinetic energy of the electron (it assumes that the emitted electron has no kinetic energy). But the thermodynamic expression assumes thermal equilibrium, so which is it? Well, at zero Kelvin there is no translational energy, so when considering $\Delta U(\pu{0 K})$ we can safely ignore any thermal contribution from the electron. So we are ignoring the thermal properties of the released electron, and then since $\Delta n =0$ (ignoring the electron) we have that $$\Delta H = \Delta U$$