Why do different gases have the same average kinetic energy at the same temperature?

What you describe is the difference between temperature and heat. If you declare that two gases have the same total kinetic energy, this implies that they have the same heat. The mass of the molecule is related to the heat capacity (for ideal gases).

Edit/answer to comment: this is mainly a mistranslation/everyday use of the word heat in my mothertongue, which probably doesn't work in english. More precisely, the total kinetic energy equals the total heat the gas could theoretically transfer to a 0K object with infinit heat capacity, assuming no phase transitions or volume change. I think this is helpfull nevertheless.

The molecules in a gas are in constant collision. Conservation of energy means that this will randomise the distribution about the mean energy level, which must therefore be the same for both gases. If this is not sufficient argument, the zeroth law of thermodynamics will ensure that the gases are in thermal equilibrium.

The thermoynamic equilibrium in a gas (here a monoatomic perfect gas as you consider only translation and ignore interactions) , as long as the temperature is fixed, is characterized by the Boltzmann distribution of velocities, stating that the density of probability in the space of velocities is proportional to $exp(-E/k_BT)$ (where $k_B$ is the Boltzmann constant), and here $E$ is simply the translational kinetic energy $E=frac12 m vec{v}^2$. An important consequence is the so called "equipartition of energy" :

$$ langle Erangle equiv frac12 m langle vec{v}^2rangle equiv frac12 m v_{rms}^2 = frac32 k_B T.$$

where the 3 comes from the 3 directions of the space.

This holds independently of the mass $m$ of the gas constituants, and is still true for a diluted diatomic gas like in air. Hence, one has actually:

$$frac12 m_1 v_{rms,1}^2 = frac12 m_2 v_{rms,2}^2=frac32 k_B T,$$

and the relation for the temperature that you give is the "thermodynamic definition of temperature". It is also the fundamental principle of the "gas themometer" used by metrologist for low temperature measurements.

Furthermore, for a perfect gas, and for a given pression and temperature, the number $N$ of molecules depends only on the volume $V$ and not on the molecular weight, and the total translational kinetic energy for this volume is the same for any gas.

Nevertheless, this does not imply that the total kinetic energy is the same, as for poly atomic gas, you have to consider not only the motion of the barycenter (as we have done above) but also the kinetic energy associated to the relative motion of the atoms in the molecules, making the laws more complex.

It would not make sense to say that the total energy has to be the same for different gases. If that were true, I could say that the total energy of 1 kg of air should be the same as the total energy of, say, 5000 kg of water vapour, which is of course absurd. Total energy is an extensive quantity: it depends on the system size. Average energy, on the other hand, is an intensive quantity, so it makes sense to compare it between systems of different size.

From kinetic theory of gases, we know that,

Pressure $(P)=frac{1}{3}frac{nM}{V}v^2$ , $ spacespacespacespacespace -(1)$

where,

$n=$ number of moles, $M=$ molar mass, $V=$ volume, $v=$ root-mean-squared velocity.

And, Kinetic Energy $(K) = frac{1}{2}mv^2 = frac{1}{2}(nM)v^2 =frac{3}{2}PV $, (from $(1)$).$spacespacespacespacespace -(2)$

But, we define absolute temperature scale as,

$frac{P}{P_{tp}}=frac{T}{273.16 K}$, where $T=$ temperature.

From here, we get, $P=P_{tp} space (frac{T}{273.16 space K})$

Using $(2)$, we get,

$K=frac{3}{2}[P_{tp} space (frac{T}{273.16 space K})]space V$

Now, $P_{tp}=$ constant, $Rightarrow$ if $V=$ constant, $Kpropto T$.

Hence, we arrive at the required expression.

You may think of it in the way that increasing temperature implies that you are supplying heat energy to the substance. As a result, for a fixed a volume, $T equiv K_{avg}$. Hope this clears your doubt!