Part One: A Globular light source emitting light in all directions.
The question says:
How far away and how large would this light source need to be, to be visually indistinguishable from how we experience the sun on Earth?
The light source would have to be rather far away to be visible from about half of the surface area of the planet at any one time, as the Sun is from the Earth.
The question doesn't describe the planet so it could be much larger or smaller than the Earth.
It is impossible for a globular light source spreading its light in all directions illuminate exactly one half of a spherical object, because the rays from the light source would have to be parallel to illuminate exactly half the spherical object, and the spherical light source would have to be infinitely far away for its beams to be parallel when they strike the spherical object.
In the case of Earth and the Sun, the Sun is about a hundred times as wide as the Earth and it's distance is over ten thousand times the diameter of the Earth.
And you can make the light source much bigger and father away, relative to the planet, than the Sun is to Earth, or make it much smaller and closer relative to the planet than the Sun is to the Earth. If you want the light source to have the same angular diameter as seen from the planet as the Sun from the Earth, you have to keep the ratio between the diameter and the distance of the light source the same as for the Sun and the Earth.
If you want a relatively small and nearby light source but want about an entire hemisphere of the planet to be illuminated at any one time there will be a limit to how small and close the light sources can be. If it get too close it will illuminate only a part of one hemisphere at a time and many parts of the planet will be in eternal darkness, which is not eh case on Earth.
So now I will have to draw a picture with words since I don't have a drawing program, and you have to draw it yourself or picture it in your mind's eye.
Picture a circle, representing your planet, with a long line though the center of the planet. Put a dot on the line representing the light source. Draw a line from the light source to graze the circumference of the circle representing the planet.
Draw a line from the point where the line from the light source grazes the circle to the center of the circle. There should be a right angle of 90 degrees at the the pint where the angle from the light sources intersects the lien frm the circumference to the center of the circle. You have now made a right angled triangle. The angle at the center of the circle representing the planet, and the angle at the light source should add up to 90 degrees, but they don't have to be equal to each other.
In order for the light source to illuminate exactly one hemisphere of the planet the angle at the center of the planet has to be exactly 90 degrees, the angle where the lien from the light source grazes the surface of the circle/planet has to be exactly 90 degrees, and the angle at the light source has to be exactly zero degrees, which is impossible in a triangle.
So the light source will be unable to illuminate exactly one hemisphere of he planet, because there will be a part of the near hemisphere in shadow where the curve of the planet blocks the light from the light source. Since you want approximately half of the planet illuminated by the light source at any one time, you need to figure out how close to a full hemisphere you want.
If, for example, you want the angle at the center of the planet to be 75 degrees, the angle at the light source will be 15 degrees.
If, for example, you want the angle at the center of the planet to be 80 degrees, the angle at the light source will be 10 degrees.
If, for example, you want the angle at the center of the planet to be 75 degrees, the angle at the light source will be 5 degrees.
If, for example, you want the angle at the center of the planet to be 89 degrees, the angle at the light source will be 1 degree.
So you could draw diagrams and measure the ration of the line between the light source and the center of the planet to the line from the center of the planet to the point where light grazes the surface.
Or you can use trigonometry to calculate the ration and see how many planetary radii the distance to the light sources is with the proportion of the surface you decide to have illuminated.
The planet is supposed to be otherwise like Earth, and so it should have a breathable atmosphere similar to Earth's. Earth's atmosphere refracts light, bending it down to illuminate areas a few degrees beyond the bulge which cuts off light in a straight line, so the planet should not be too unlike Earth if the line from the center to the edge where light grazes the surface is just a few degrees.
Part Two: A disc like light emitter with parallel rays.
Of course the light source doesn't have to be a spherical light source emitting light in all directions. It could emit light from a disc aimed at the spherical world, and that light could be parallel like laser light.
Because the light rays would be parallel in that case, the light emitting disc would have to be at least as large as the diameter of the spherical planet, in order for one entire hemisphere to illuminated.
If the light emitting disc pointed at the planet has the the diameter of the planet, and also has an angular diameter of about 0.5 degree, at the distance to the planet, my rough calculations go like this:
One half of a degree is one 720th of a full circle of 360 degrees. The circumference of a circle is 2 times pi times the radius of the circle. Using 3.14159 as a good enough value of pi, The radius of a circle is equal to 57.295 degrees of the circumference. Since the light disc is equal to the planet's diameter and to 0.5 degees of the circumference, the radius of the circle and the distance between the light disc and the planet is equal to 114.5916 times the diameter of the light disc and of the planet.
If the planet has a diameter of 6,371 kilometers, the mean diameter of Earth, the distance to the light disc will be about 730,063.4 kilometers.
If the light disc is more than a few tens of thousands of kilometers closer than the calculated distance, its disc will appear to be larger than the Sun appears from Earth, and if it is more than a few tens of thousands of kilometers farther than the calculated distance, it will appear noticeably smaller than the Sun appears from Earth. And if the light disc is farther from the planet than the calculated distance its parallel rays will illuminate less than one full hemisphere of the planet at a time.
I would be grateful for anyone who improved by answer with diagrams and/or trigonometric calculations.