These Wikipedia images should make it clear:
You want to draw an image which is distorted like the right one. Your VR system will map it onto a sphere like in the left and the system user watches it sitting in the midpoint of the sphere.
You "simply" take a rectangular area which is 2 times as wide as it's high. You want obviously draw space dust, stars and maybe some imagined planets, too.
Put the items into the rectangle, but you must decide the right scaling. Vertically there's needed no scaling. But horizontally you must stretch the items. The more stretching the further the item is from the equator. Single points, the north and south poles will fill alone the top and the bottom edges of the rectangle. Small items on the equator could be drawn as is without error.
In theory you can draw without stretching and with mathematical transform you can create the stretched image. The green area shows how much room you have:
If you insert the items into green area, cut it to small horizontal slices and stretch every green part as wide as the rectangle, you have the wanted equirectangular image.
The exact form of the edge of the green area is a half cycle of cosine curve (it's the same as sine curve, but shifted)
If you do the stretching to both directions, the job can seem subjectively easier:
In both cases the total stretching is the same (=nothing at the equator, one point to whole image width at the poles).
The practical drawing onto the green area is extremely difficult, because only very small items do not get distorted.
Drawing can be helped with lines which will become horizontal and vertical in the stretched image and present meridians and parallels on the globe map.
Parallels in unstretched image are the same as in the stretched image, but meridians are differently scaled cosine curves:
The stretching is possible in programs which have mapping with custom equations. The mapping takes content to point (x,y) in the stretched image from unstretched image point (x',y'), where y' =the same y and x' =(W-1)/2 + (x-(W-1)/2)cos(Pi(y-(H-1)/2)/(H-1)).
W and H are the image size in pixels, the coordinates are assumed to have ranges from 0 to (W-1) and (H-1). This is for bidirectional stretching, one directional stretching has nearly same formula , only replace (W-1)/2 with (W-1).
This is the stretched image and the stretching dialog in Affinity Photo:
The numbers reflect the dimensions W=3000, H=1500. As a photo this could be quite high resolution, but for this it's too coarse - as you see, polar areas are fuzzy. The image should contain sharp rectangles.
Placed on the sphere the result is a little cleaner due the pinched polar areas, but still far from the ideal:
I tested also actual drawing. Directly as stretched it was hopeless, nothing became recognizable near poles and the rest was (at best) laughably distorted, when the result was projected on a sphere.
Filling the green area before stretching worked marginally better, if the meridians and parallels were in place for reference, but in practice it was also hopelessly non-productive.
ADD: User Gerald Falla has told in his answer how it should be done. We can try it.
Actually Affinity Photo has live equirectangular mapping. One makes an empty drawing say 3000 px wide and 1500 px high, fills it (if needed) with squares for easier orientation and goes to layer live equirectangular projection mode. There he can paint the shapes, but for feedback he watches the sphere from the centerpoint.
There should be no problem to get the shapes drawn undistorted, if the person can draw well on normal plane. The drawing generates under the hood the right equirectangular image.
At first we have 3000x1500 px bitmap image with rectangular grating. The equator and the zero-meridian are red.
Of course it sits on a sphere:
Let's goto the live equirectangular layer projection and pan the north pole into the viewport. Then draw with mouse something:
When the layer live projection was closed, the equirectangular image was shown:
I bet only Leonardo Da Vinci or Archimedes could see what it is. On sphere it's this (as flipped):