In ideal gas model, temperature is the measure of average kinetic energy of the gas molecules.
In the kinetic theory of gases random motion is assumed before deriving anything.
If by some means the gas particles are accelerated to a very high speed in one direction, KE certainly increased, can we say the gas becomes hotter? Do we need to distinguish the random vibration KE and KE in one direction?
The temperature is still defined by the random motion, subtracting the extra energy imposed . This is answered simply by the first part of @LDC3 's answer. Does your hot coffee boil in the cup in an airplane?
Furthermore, if we accelerate a block of metal with ultrasonic vibrator so that the metal is vibrating in very high speed with cyclic motion, can we say the metal is hot when it is moving but suddenly become much cooler when the vibration stop?
This is more complicated, because vibrations may excite internal degrees of freedom and raise the average kinetic energy for that degree of freedom. It would then take time to reach a thermal equilibrium with the surroundings after the vibrations stop. If one supposes that this does not happen, then the answer is the same as for the first part, it is the random motions of the degrees of freedom that define the kinetic energy which is connected to the definitions of temperature. So no heat will be induced by the vibrations.
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" :
$$ \langle E\rangle \equiv \frac12 m \langle \vec{v}^2\rangle \equiv \frac12 m v_{rms}^2 = \frac32 k_B T.$$
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.
Best Answer
The main difference between gases and solids is indeed that in the latter, interactions between constituents are not negligible.
However, i do not think that your approach is valid, because the right result should be (using your notations) $K_s = K_g$.
Here is an approximate explanation :
The situation of a solid can be modelled this way : the interaction between two particles of this solid is described by a potential $U$ that is a function of their relative position. For reasonably low temperatures, the particles will remain close to their equilibrium position and oscillate thermally around it. For small deviations around this position, the position can be approximated to be $$U(\vec{x}) \simeq \frac{1}{2} k \vec{x}^2$$ Where $x$ is the deviation of a particle from its equilibrium position.
For oscillators, the energy is equally distributed between potential and kinetic energy so that : $$\frac{1}{2}m\langle \vec{v}^2 \rangle = \frac{1}{2}k \langle \vec{x}^2\rangle$$
As stated by the equipartition theorem, to each degree of freedom of the system corresponds an average energy $\frac{1}{2} k_B T$. So, just like in gases, the average kinetic energy is still given by : $$\frac{1}{2}m\langle \vec{v}^2 \rangle = \frac{3}{2} k_B T$$
What really differs is that when interactions aren't negligible anymore, the total energy is the sum of the kinetic energy and the potential energy which is on average (using the two previous equations) $$\frac{3}{2} k_B T + \frac{3}{2} k_B T = 3 k_B T$$ This might sound surprising that molecules can go as fast in a solid as in a gas, but note that in a solid they don't go further than $\langle \vec{x} \rangle$ on average and their high kinetic energy is due to their oscillation frequency $\omega = \sqrt{\frac{k}{m}}$.
Side note
Since the average thermal energy for one particle of the solid is $3k_B T$, the average energy for $N$ solids is $3Nk_B$. As a result the heat capacity of a solid in $J/K/mol$ should approximately be around $3R$, and this result, known as the Dulong–Petit Law, has some experimental support, giving some credit to this approach.