edit im retarded
ok after thinking some more. if A is invertible, then it is row equivalent to I. you would think adding two row equivalent matrices would be row equivalent so A would be row equivalent to (A+I), but theres actually an exception here...when A has a row which is equal to the same row in -I, then adding that to I will mean that to get to (A+I) from A, you must perform the row operation of adding the negative of a row to itself, which is the same as multiplying by 0. this isnt a valid row operation so they aren't row equivalent. however this is the only exception.
so this means that either (A+I) is invertible, or it has at least one row of all 0's.
plugging this back into the equation XA(A + I) = A + I doesn't really help though. if all of the rows are 0 (A=-I), then it obviously solves for all X, but if only some rows are 0, I can't think of anything that would mean, especially since A+I is on the right side of the multiplication. if it was on the left side then that would mean an entire row of the product would also be 0 which would be equal to A+I, but on the right side that doesnt seem to necessarily be the case. you can use distributive property on other side to prove (A+I)A has the same zero rows, but you’re still left multiplying that by X so it doesnt help.