Since potential energy is considered non-existent in ideal gases,
kinetic energy = internal energy. However the 2 formulae lead to
different results. What is the reason for this, what am I missing?
While your first statement is true, please note that you're wrong in saying $PV = U$.
As I show below the correct relationship betweeen $PV$ and $U$ comes with a factor of $2/3$.
We first note that internal energy ($U$) of an ideal gas is only kinetic energy ($K$), as there is no potential of interaction. Therefore, firstly we have,
$$
U = K
$$
Now let us see from kinetic theory of gases, how we can calculate pressure ($P$) from kinetic energy ($K$). Physically, pressure is the force per unit area exerted by the gas atoms/molecules on the walls of the container.
Consider a gas of a large number $N$ of molecules, each of mass $m$, enclosed in a cube of volume $V = L^3$. When a gas molecule collides with the wall of the container perpendicular to the $x$ axis and bounces off in the opposite direction with the same speed (an elastic collision), the change in momentum is given by:
$$
\Delta p = p_i - p_f = p_i - (-p_i) = 2mv_x
$$
where $p_i$ represents the initial momentum of the molecule and $p_f$ represents the final momentum after the collision. Here $m$ is mass and $v_x$ the velocity in $x$-direction.
Let us say that the particle impacts one specific side wall only once during the time interval $\Delta t$, therefore,
$$
\Delta t=\frac{2L}{v_x}
$$
Therefore, the force exerted on the wall due to $N$ particles is,
$$
F = N\frac{\Delta p}{\Delta t} = N\frac{m v_x^2}{L}
$$
Assuming the motion of particles is isotropic, we get, mean squared speed $\left< v^2 \right> = \left< v_x^2 \right>/3$, and hence the force can be written as,
$$
F = \frac{Nm \left< v^2 \right>}{3 L}
$$
which leads us to the expression for pressure, force on an area $L^2$
$$
P = \frac{F}{L^2} = N\frac{m\left< v^2\right>}{3L^3}=\frac{2}{3}N\left(\frac{m\left< v^2 \right>}{2}\right) \frac{1}{V} = \frac{2}{3} \frac{K}{V} = \frac{2}{3} \frac{U}{V} \\
\implies PV = \frac{2}{3} U
$$
Now, if you subsititute the values for average kinetic energy per molecule, you'll get the ideal gas equation.
