Following is the way I figured out the answer. I hope this helps to simplify things a bit.
Let us first write the resultant force on the box in the frame in which the incline is at rest.
$$\mathbf{F_r} = mg \sin \theta \cos \theta \mathbf{\hat{x}} - mg \sin ^2 \theta \mathbf{\hat{y}}$$
This resultant force is the same in all inertial frames. Hence, acceleration is same in both the frames.$$\mathbf{a_r} = g \sin \theta \cos \theta \mathbf{\hat{x}} - g \sin ^2 \theta \mathbf{\hat{y}}$$
Consider first the situation as seen in the frame in which the incline is at rest.
Initial velocity $\mathbf{u_0} =0$ Considering the point of release as the origin, initial position $x_0 = 0;\ y_0 =0$. Solving Newton's laws we get
$$x = \frac{1}{2} g \sin \theta \cos \theta \ t^2 \space \ \ \ \ y= -\frac{1}{2}g \sin^2\theta \ t^2 \\ v_x = g\sin\theta\cos\theta\ t \ \ \ \ \ v_y =-g\sin^2\theta \ t$$
Work done is just
$$W = \int\ \mathbf{F_r \cdot dr} \ = -\Delta PE = \frac{1}{2}mg^2\sin^2\theta\cos^2\theta\ t^2\ + \frac{1}{2}mg^2\sin^4\theta\ t^2\\=\frac{1}{2}mv_x^2\ +\ \frac{1}{2}mv_y^2\ = \Delta KE$$
And we have energy conservation $\ \Delta KE\ +\ \Delta PE\ =\ 0 $
Now to the situation which OP has asked. Let the time taken for the box to reach the bottom of the incline be $t_f$ (same in both frames). The relative velocity between the frames as posed by OP is $\mathbf{v_f} =g\sin\theta\cos\theta t_f \mathbf{\hat{x}}$. Let the origins coincide at $t=0$. We then have $x_0=0;\ y_0 =0$ and $\mathbf{u_0}= -\mathbf{v_f}$. Acceleration remains the same in both the frames. Solving Newton's laws we get
$$x = -g\sin\theta\cos\theta t_f t+\frac{1}{2} g \sin \theta \cos \theta \ t^2 \space \ \ \ \ y= -\frac{1}{2}g \sin^2\theta \ t^2 \\ v_x = -g\sin\theta\cos\theta t_f+g\sin\theta\cos\theta\ t \ \ \ \ \ v_y =-g\sin^2\theta \ t$$
Work done is just
$$W = \int\ \mathbf{F_r \cdot dr} \ = -\Delta PE = - mg^2\sin^2\theta\cos^2\theta t_f t+\frac{1}{2}mg^2\sin^2\theta\cos^2\theta\ t^2\ + \frac{1}{2}mg^2\sin^4\theta\ t^2\\$$ At $t=t_f$
$$W = -\Delta PE = - mg^2\sin^2\theta\cos^2\theta t_f^2+\frac{1}{2}mg^2\sin^2\theta\cos^2\theta\ t_f^2\ + \frac{1}{2}mg^2\sin^4\theta\ t_f^2\\ =-\frac{1}{2}mg^2\sin^2\theta\cos^2\theta\ t_f^2\ + \frac{1}{2}mg^2\sin^4\theta\ t_f^2\\ =-\frac{1}{2}mv_x^2(t=0)\ +\ \frac{1}{2}mv_y^2(t=t_f)\ = \Delta KE$$
Since $v_x(t=t_f)=0;\ v_y(t=0)=0$.
And we have energy conservation $\ \Delta KE\ +\ \Delta PE\ =\ 0 \ $!!!
Excuse me.. What was the problem again?
Well... the apparent catch was in posing the problem. While posing the problem OP has used $\Delta KE$ and $\Delta PE$ measured in different frames to verify Law of conservation of energy. The problem considers only change in $\Delta KE$ and not change in $\Delta PE$ when we change from the stationary frame to the moving frame. It is to be noted that both $\Delta KE$ and $\Delta PE$ change. Since the frames will have relative motion, the velocities of the particles change according to velocity addition formula and kinetic energy changes. Lets say in the initial frame for a given particle $$\Delta KE = \frac{1}{2} m (u_f^2 -u_i^2)$$If we now wish to see it in a frame with a relative velocity $-v$, the velocities get transformed into $u_f + v$ and $u_i +v$. Thus change in kinetic energy changes to $$\Delta KE = \frac{1}{2} m (u_f^2 -u_i^2) + mv(u_f-u_i)$$ To compensate for this gain in $\Delta KE$, Work done has to change to keep conservation of energy intact. We can show that it is exactly the same as the gain in $\Delta KE$. On changing the frame,$$\mathbf{dr}\longrightarrow\mathbf{dr}+\mathbf{v}dt$$ In the new frame,$$W=\int\mathbf{F}\cdot\mathbf{dr}+\int\mathbf{F}\cdot\mathbf{v}dt = \int\mathbf{F}\cdot\mathbf{dr}+\int m\dfrac{d\mathbf{u}}{dt}\cdot\mathbf{v}dt =\int\mathbf{F}\cdot\mathbf{dr}+mv(u_f-u_i)$$
Hence, both $\Delta KE$ and $\Delta PE$ have to be frame dependent.
Note:-
1) The above discussion assumes that the forces in the problem are time independent and conservative. It is only then that we can define $\Delta PE$ as $W = -\Delta PE$. Which is not true if it were time dependent or non-conservative. In which case one cannot define potential energy. We will then be proving work kinetic energy theorem instead of $\Delta KE+\Delta PE=0$.
2) One might argue that the work done here was by the constraint force. Frankly, as I have proved, it should be true for any force. In this case it happens to be generic to the problem that the component of the resultant force in the direction of relative motion (x direction) happens to be contributed only by the constraint force.
Consider for instance the following situation. A $2 kg$ weight is made to fall freely under the influence of gravity from rest for a distance of $5 m$. Using $g=-10 m/s^2$ we find that the velocity after $5m$ is $-10 m/s$.$\Delta KE = 100 kgm^2/s^2$. Now let us observe this from a frame moving downward with a velocity $-5 m/s$. In this frame, initial velocity is $5m/s$ and initial kinetic energy is $25 kgm^2/s^2$. After falling for $5m$, the velocity of the particle will be $-5m/s$ and final kinetic energy will be $25 kgm^2/s^2$.$\Delta KE = 0 kgm^2/s^2$. Oh... there was no change in potential energy even as the particle fell for $5 m$ !!!
The resolution for this goes in the similar lines as in the situation above. It is just for gravity in this case.
3) It is to be noted that the change in work done on changing frames depends only on the initial and final velocities of the particle irrespective of what it has gone through in between!!
4) Throughout the calculation, I have considered real conditions of experiment i.e. a typical box and typical inclined plane on earth whose masses are small compared to earth. Acceleration due to gravity is constant. If people wish to consider this as an approximation, people are welcome to solve the problem in all detail to their satisfaction. But the result I have proved above is a very profound truth independent of the case at hand.