Whether there is vacuum or dielectric, there is only one dielectric constant difference in the equation, which has no special significance (dielectric polarization leads to the weakening of electric field intensity). Only the electrostatic field in vacuum is discussed here.
Gaussian equation describes that there are many point charges on the closed surface in vacuum, so the electric field intensity flux excited by these point charges is a certain value. Electric field intensity flux = product of electric field intensity e and unit area of relevant closed surface, and integral of the whole closed surface. When the closed surface is spherical, the classical formula of electric field intensity excited by point charge is obtained. Electrostatic field excited by electrostatic charge is the source of electric field intensity flux of electrostatic field. Therefore, the classic field is an active field.
The point charges outside the closed surface all enter at one end and exit at the other end, and the flux is zero, so the external charge can be omitted when calculating the electric field intensity flux of a certain surface. However, Gauss theorem only discusses the flux problem, and calculates the excitation field strength of point charge through transformation. When calculating the electric field intensity excited by multiple charges at a certain point, the intensity of superimposed electric field must be calculated independently by Gauss theorem.
Conclusion: Your first sentence is wrong. Gauss theorem does not explain that the electric field intensity e is determined by both internal and external charges. Gauss theorem only says that when calculating electric field intensity flux, it is only related to internal charge. To calculate e, Gauss theorem and superposition operation should be used respectively. Gauss theorem is of great value, far from the most basic application of calculating electric field intensity.