Two points above an observer do not maintain their apparent angular displacement no matter how far they are above the observer.

If we imagine that the image you provided was a three dimensional scene starting with those stars at the same altitude as the observer, if we then increase the altitude of the stars over the observer nearest the North Pole the angle the observer sees between the two top and bottom stars would become less and less as the stars get further and further from the observer.

If, instead, we imagine that the previous image I provided is three dimensionally sliced through the center with a copy of itself on other axis like a + when viewed from overhead, creating a symmetrical three dimensional scene with four stars instead of two, we can see that the consistent angles would also work on the other axis.

You're absolutely right in the sense that the angle we're interested in is the apparent azimuth as far as the viewer is concerned - the azimuth angle is tilted up at the elevation angle, as if the observer was making an azimuth measurement. So yes, my diagram is something of a simplification, in that the two pairs of red lines aren't precisely comparable, but the point I'm making is that the difference is enormous, and correcting for elevation doesn't fix the problem.

Let's take an example, with one viewer at the North Pole observing a star that is 5 degrees away from the pole, so roughly 300nm laterally, and one observer at 30 degrees north, so 3600nm away. The calculation for our polar observer is simple - it's 5 degrees, whichever way he looks, and wherever the star is on its circular track. I've kept the example deliberately small so we don't stray into needing to make big corrections for EA...difficult to do as you haven't actually got a formula for it! I hope you agree that the difference from a straight line is pretty negligible at 5 degrees.

Our more southern observer requires some maths. The position is 3600nm horizontally, and displaced 5400nm vertically according to your updated number. Pythagoras gives us a direct viewing distance for Polaris of around 6500nm. The other star is roughly 300nm displaced from the pole at the same altitude (we're assuming it's at its maximum azimuthal displacement, 3 or 9 o'clock around Polaris with respect to the ground/observer). So now our 6500nm viewing line becomes the adjacent side of a new right angled triangle, the 300nm becomes the opposite. Trigonometry gives an azimuth angle of around 2.6 degrees for our distant observer (tan 2.6 = 300/6500), so roughly half the azimuth angle for our polar observer.

That's a massive difference - that means the distant observer would see elliptical movement, with the ellipse roughly twice as high as its width. That is not what we observe, is it? Do check my maths, of course...it's late.