well...if you were to dissect this into a real physics problem, you'll have the following:
"If everyone aimed a laser pointer at the Moon at once, would it change color?"
1) the moon is only visible to 50% of the earth's surface at any one time
2) on that 50% of the surface, only 30% is land on average.
3) on that land portion, there'd a totally undeterminable number of human population. but if we assume a uniform distribution of human population of land (totally idiotic) across that land mass, we could at least have something. so 7bil population uniformly spread across perfectly spherical surface of earth. ~3.5billion can see the moon purely from line of sight, not taking into account surface toplogy, cloud cover, etc etc.
~3.5billion * 5mw laser pointers = 17.5 megawatt output. but this is only output, just the tip of the iceberg.
since the population is spread uniformly across the applicable surface, the laser that's shining directly perpendicular to the surface of the earth is going to be closer to the moon than the laser that's shining tangent to the surface of the earth. within the earth atmosphere, the laser light will be refracted and diminished related to the distance traveled. additionally, temperature, humidity, pollution particulate matter content will all contribute to rob the laser beam of its power.
even once the laser beam has exited the atmosphere, it still has to contend with space dust, although minutely. once it reaches the moon, once again, lunar atmosphere. once it reaches surface of the moon, how lunar rocks/dust reflect the laser beam will need to be taken account of. and again on the trip back to the earth, lunar atmo, space dust, earth atmo, and then finally perceived brightness on observer's retina.
there's a whole lot of data and number crunching to be had. but it's just easier to use an existing item as a base.
there are lunar observation reflectors are planted on the surface of the moon by apollo missions and by soviet missions too. according to wiki
Lunar Laser Ranging experiment - Wikipedia, the free encyclopedia, "The reflected light is too weak to be seen with the human eye: out of 10^17 photons aimed at the reflector, only one will be received back on Earth every few seconds, even under good conditions."
so given a return ratio of 1/10^17, and an output power of 17.5megawatt, you will only get to see 0.00000000017 watts returned. so no....you won't see a diff hehe