GetWindowRect returns a size including "invisible" borders

How can I convince Windows to let me work with the real window coordinates?

You are already working with the real coordinates. Windows10 has simply chosen to hide the borders from your eyes. But nonetheless they are still there. Mousing past the edges of the window, your cursor will change to the resizing cursor, meaning that its still actually over the window.

If you want your eyes to match what Windows is telling you, you could try exposing those borders so that they are visible again, using the Aero Lite theme:

http://winaero.com/blog/enable-the-hidden-aero-lite-theme-in-windows-10/

Windows 10 has thin invisible borders on left, right, and bottom, it is used to grip the mouse for resizing. The borders might look like this: 7,0,7,7 (left, top, right, bottom)

When you call SetWindowPos to put the window at this coordinates:
0, 0, 1280, 1024

The window will pick those exact coordinates, and GetWindowRect will return the same coordinates. But visually, the window appears to be here:
7, 0, 1273, 1017

You can fool the window and tell it to go here instead:
-7, 0, 1287, 1031

To do that, we get Windows 10 border thickness:

RECT rect, frame;
GetWindowRect(hwnd, &rect);
DwmGetWindowAttribute(hwnd, DWMWA_EXTENDED_FRAME_BOUNDS, &frame, sizeof(RECT));

//rect should be `0, 0, 1280, 1024`
//frame should be `7, 0, 1273, 1017`

RECT border;
border.left = frame.left - rect.left;
border.top = frame.top - rect.top;
border.right = rect.right - frame.right;
border.bottom = rect.bottom - frame.bottom;

//border should be `7, 0, 7, 7`

Then offset the rectangle like so:

rect.left -= border.left;
rect.top -= border.top;
rect.right += border.left + border.right;
rect.bottom += border.top + border.bottom;

//new rect should be `-7, 0, 1287, 1031`

Unless there is a simpler solution!