You have successfully discovered that the kinetic energy depends on the reference frame.
That is actually true. What is amazing, however, is that while the value of the energy is frame DEpendent, once you've chosen a reference frame, the law of conservation of energy itself is NOT reference frame-dependent -- every reference frame will observe a constant energy, even if the exact number they measure is different. So, when you balance your conservation of energy equation in the two frames, you'll find different numbers for the total energy, but you will also see that the energy before and after an elastic collision will be that same number.
So, let's derive the conservation of energy in two reference frames. I'm going to model an elastic collision between two particles. In the first reference frame, I am going to assume that the second particle is stationary, and we have:
$$\begin{align}
\frac{1}{2}m_{1}v_{i}^{2} + \frac{1}{2}m_{2}0^{2} &= \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2}\\
m_{1}v_{i}^2 &= m_{1}v_{1}^{2} + m_{2}v_{2}^{2}
\end{align}$$
to save myself time and energy, I'm going to call $\frac{m_{2}}{m_{1}} = R$, and we have:
$$v_{i}^{2} = v_{1}^{2} + Rv_{2}^{2}$$
Now, what happens if we shift to a different reference frame, moving to the right with speed v? This is essentially the same thing as subtracting $v$ from all of these terms. We thus have:
$$\begin{align}
(v_{i}-v)^{2} + R(-v)^{2} &= (v_{1}-v)^{2} + R(v_{2}-v)^{2}\\
v_{i}^{2} -2v_{i}v + v^{2} + Rv^{2} &= v_{1}^{2} - 2 vv_{1} + v^{2} + Rv_{2}^{2}-2Rv_{2}v + Rv^{2}\\
v_{i}^{2} -2v_{i}v &= v_{1}^{2}- 2vv_{1} + Rv_{2}^{2}-2Rv_{2}v\\
v_{i}^{2} &= v_{1}^{2} + Rv_{2}^{2} + 2v(v_{i} - v_{1} - R v_{2})
\end{align}$$
So, what gives? It looks like the first equation, except we have this extra $2v(v_{i} - v_{1} - R v_{2})$ term? Well, remember that momentum has to be conserved too. In our first frame, we have the conservation of momentum equation (remember that the second particle has initial velocity zero:
$$\begin{align}
m_{1}v_{i} + m_{2}(0) &= m_{1}v_{1} + m_{2}v_{2}\\
v_{i} &= v_{1} + Rv_{2}\\
v_{i} - v_{1} - Rv_{2} &=0
\end{align}$$
And there you go! If momentum is conserved in our first frame, then apparently energy is conserved in all frames!
I didn't redo your calculations and I assume that they are correct, which actually doesn't play any role in what I'll describe now. Notice that in the second scenario the 2kg ball will inevitably start to move. By keeping it still you change the reference frame one more time, which invalidates the use of conservation laws.
You cannot use the conservation of energy or momentum in two reference frames$^*$. It is pretty straight forward to see why this must be the case. Assume there is a ball with mass $m$ and from the reference frame of the stationary observer it is moving with velocity $v$. Clearly the ball has a kinetic energy $mv^2/2$ and a momentum $mv$. However from the reference frame of the ball, it is not moving therefore its momentum and energy is zero. If you were to use the conservation of energy or momentum, you would see that it is violated. Thus you conclude that you cannot use the conservation laws for two different reference frames.
I am not sure I understand what your last question is.
$^*$I am assuming that you don't know about momentum-energy four vector, which combines these two and is in fact conserved.
Best Answer
Conservation of energy refers to systems looked from the same reference frame, it does not make sense to require that energy of the same system to be the same in different reference frames. As a consequence of time translational symmetry, energy conservation is usually true unless we drive the system externally which may break this symmetry. Similarly, momentum conservation is a consequence of space translational symmetry.
The (invariant) mass $m$ is the same in all inertial reference frames, on the other hand, energy $E$ and momentum $p$ are connected through the famous equation
\begin{equation} E^2=(pc)^2+(mc^2)^2 \end{equation} where $c$ is the speed of light. This equation is valid in any inertial reference frame, to go from one frame to another, one has to do Lorentz transformation of both energy and momentum, and it turns out the final result is that the changes in energy and momentum compensate each other and validate this equation in every frame.
For the example you gave, if there is only that ball in the universe, in reference frame A, it cannot stop by momentum conservation. If it stops, you have to exert an external force, which may explicitly change the energy of this ball even in reference frame A. Then from frame B, roughly speaking, you exert a force (you may want to work out the transformation of the force between these two frames) to the left direction and the ball gains energy, so nothing is wrong.