Is mechanical energy conserved in all Inertial frames? (Newtonian Mechanics)

TLDR: In Newtonian mechanics the change in potential energy is frame independent and, in the absence of dissipation, total mechanical energy is conserved in all frames. The confusion you are encountering is not due to any problem with the definition of potential energy or mechanical energy, it is due to not considering the total mechanical energy.

Unfortunately, I must disagree with the other answer. This is a confusing topic, but it is incorrect that:

the work done by gravity (which is now also the change in potential energy)

The change in potential energy remains $mgDelta h$ (where $h$ is the separation) although the work done by gravity does not. The potential energy must remain the same or you could create a perpetual motion machine by for example compressing a spring at rest and then releasing it while moving.

Let's investigate this more fully. For small changes in separation, $Delta h$ between the ball and the earth the change in gravitational potential energy is $Delta U = mgDelta h$. Note that this potential energy does not belong to the ball itself, but rather to the ball-earth system. In other words, $Delta h$ is not a property of the ball only, but a property of the ball-earth system, the configuration of the two. That means that if you are considering the gravitational potential energy then you must consider the entire system of both the ball and the earth. The other analyses omitted the earth.

The mechanical energy is the sum of the kinetic energy of the Earth, the kinetic energy of the ball, and the potential energy, $E=T_E+T_B+U$, where $T_E=frac{1}{2}M V^2$, $T_B=frac{1}{2} m v^2$, and $U=mgh$. In a frame where the earth and ball are initially moving at $u$ it can be shown that $$T_E(t)=frac{1}{2}Mu^2+mgut+frac{1}{2}frac{m^2}{M}g^2 t^2$$ $$T_B(t)=frac{1}{2}mu^2-mgut+frac{1}{2}mg^2 t^2$$ $$U(t)=mgh_0 - frac{1}{2}left(frac{m^2}{M}+mright)g^2 t^2$$ $$E(t)=frac{1}{2}(m+M)u^2+mgh_0=const.$$ So the total mechanical energy $E=T_E+T_B+U$ is conserved, but $T_B+U$ is not.

Given this, why is it that we can get away with using only $T_B+U$ in the frame where $u=0$? Note, if $mll M$ then the last term of $T_E$ drops out and it simplifies to $$T_E=frac{1}{2}Mu^2+mgut$$ This is non-zero except in the case where $u=0$ and more importantly it changes over time if $une 0$. So we can get away with the "cheat" because the Earth is much more massive than the ball and so that leads to $T_E=0=const.$ but only in the frame where $u=0$.