Short answer:
The resistor does not actually cause a "voltage drop" -- at least not directly. The potential difference across the resistor is better understood as a natural consequence of the fact that nonzero electromagnetic fields must be present within the resistor in order for a current to flow through it.
Long answer:
Consider a charge-neutral metal conductor that contains free electrons, electrons that are not bound to the atoms in the metal lattice and are therefore allowed to move freely around the material. Let the lengthwise dimension of the conductor be denoted as $L$ and the cross-section area be denoted as $A$, and for simplicity, let both of these parameters be constant. In equilibrium, the free electrons experience thermal motion which causes them to collide with the atomic lattice that makes up the bulk of the material, and the overall thermal motion of the free electrons is random such that there is no bias for a given electron to move in a specific direction. Thus the average velocity of any given free electron is $0$.
However, applying a constant $\vec E$-field in the direction parallel to $L$ will introduce a force that accelerates the electrons in the direction opposite of $\vec E$. Now that the electrons are in a non-equilibrium state, the rate of acceleration for the electrons will decrease due to collisions with the atomic lattice and other thermal effects until an equilibrium state is achieved in which the average velocity of any given electron is constant and non-zero in the direction opposite of $\vec E$. This average velocity is referred as the drift velocity of the free electrons in the metal, and the drift of these electrons is what is responsible for the macroscopic observation of a positive current $\vec I$ flowing in the direction of $\vec E$.
What I have just described is the conceptual framework for what is known as the Drude model, a classical model of electrical conduction in metals and other conductive materials based on the principles of classical electromagnetism and kinetic theory. There are more generalized and modern variations of this model that incorporate our modern understanding of electromagnetic phenomena, but this model turns out to be sufficient for giving theoretical justification for Ohm's Law, which in vector form is:
$$\vec E = \rho \vec J$$
where $\vec E$ is the applied electric field within the conductor, $\vec J = \frac{\vec I}{A}$ is the current density (the amount of current per area flowing through the cross-section area of the conductor), and $\rho$ is the resistivity of the conductor, an intrinsic property of the material.
Now consider one of the free electrons moving along the length of the metal. The amount of work per unit charge $W$ performed by the $\vec E$-field on the electron is defined as the line integral of $\vec E$ along the path $C$ that the electron takes through the metal,
$$W = \int_C \vec E \cdot d \vec l$$
Putting together this expression for work and Ohm's Law, we end up with
$$\int_C \vec E \cdot d \vec l = \int_C \rho \frac{\vec I}{A} \cdot d \vec l$$
But the left-hand side of this equation is exactly the definition of an electromotive force $V_{EMF}$ -- the amount of work per unit charge performed by the $\vec E$-field along the path that the electron takes through the metal -- and by assuming the electron moves with constant drift velocity $\vec v_d$, the path $C$ the electron takes through the metal can be assumed to be a straight line with length L, and the right-hand side of the equation reduces to
$$\int_C \rho \frac{\vec I}{A} \cdot d \vec l = \rho \frac{I}{A} L$$
Thus,
$$V_{EMF} = IR$$
where
$$R \equiv \frac{\rho L}{A}$$
is the resistance of the metal conductor with length $L$ and cross-section area $A$.
