Does the existence of electrons validate the integral form of electric fields? For an arbitrary charged object, it seems to be the case that we express it as a continuous sum (sum on the reals/integral) of point charges $dq$ that have a canonical Coulomb's law force.
That is to say, for an arbitrary charged object, we split it up into tiny $dq$'s (located at $\vec r'$, with the force exerted on reference point $P$ at $\vec r$ by them equal to..
$\text{let} \ \vec r - \vec r' = \vec \zeta$
$$F_{dq} = k \ dq \ \frac{\vec{\zeta}}{\zeta^2}$$
Implying..
$$\vec E = k \int \frac{1}{\zeta^2} \vec \zeta dq$$
But why do we assume that $dq$ exhibits the form $F_{dq}$? It's almost like there's a fundamentally point-like charged particle composing all charged objects.. aha! Electrons. But wasn't this theory established independent of electrons? How could we justify them without electrons? Do we need to? Is that even the justification for it? Why are we allowed to assume all charged objects are made of infinitesimal point charges and do electrons have anything to do with it?
 A: It was experimentally verified that assuming a continuous charge distribution was a good approximation for most, if not all, of the macroscopic electric phenomenon. A continuous charge distribution is no equivalent to a distribution of charged point particles. Assuming a continuous charge distribution can actually be thought of as in favor of the idea of continous matter distributions, which is the opposite of the idea of discreteness needed for the electron point particle.
Although it is true that electrons are point particles, if you want to rigorously deal with a collection of point charges you should not use an integral, because it is a "summation" assuming infinitely small distance between points, and that is not the case with the distribution of electrons on objects.
A: In the same sense that we can consider water to be continuous even if it is made from individual molecules, we can very often think of charges in a distribution as being continuous.  The best discussion of this is in Purcell's E&M book.
Although mathematically $dq$ is an infinitesimal amount of charge (which could numerically be less than $1.6\times 10^{-19}C$), in practice it is a sufficiently large number of point charges, with a total net charge still small enough to avoid complications due to the discreteness of the charges.  Basically, in the case of a volume charge distribution for instance, $dq_s=\rho dv$ and the volume element $dv$ is small relative to the size of our system, but still large enough to contain very many electrons (or protons, or atoms). 
In other words, we slightly abuse notation and use $dq$ rather than the strictly more correct  $\Delta q$, and compute the field as
$$
\vec E\sim \sum \frac{\Delta q_s}{\zeta_s^2}\hat \zeta_s
\approx \int \frac{dq_s}{\zeta_s^2}\hat \zeta_s \, . \tag{1}
$$
This is the converse of the usual mathematical procedure where the integral is approximated by a (Riemann) sum. 
This depends quite critically any macroscopic amount of charge being a huge number of elementary charges.  
One cannot always pass from the sum to the integral: for instance, in the classic problem of calculating the potential energy of any one charges on a line
of alternating charges each separated from its neighbour by the same distance.  In this case, it is not possible to properly define a macroscopic linear charge density and one must evaluate a sum, not an integral.
In terms of the water analogy, we have no problem writing $dm$ for an infinitesimal amount of water, even if  numerically it could be less than the mass of one molecule.   To compute the amount of water flowing through a pipe, one would normally consider $dm$ to be continuous rather than discrete.  
A: If I understand the question correctly it is about the impact on the field of the difference between a continuous and a discrete charge dustribution. This difference manifests itself as shot noise. 
