Why doesn't Current through a resistor decrease with voltage drop?

The correct understanding is "voltage drops are the push" rather than "voltage is the push".

"Voltage drop" means that the potential on one side of the resistor is lower than the potential on the other side. More importantly, positive charge carriers (conventional current) move from high to low potential (the electric force points from high to low potential for positive charges). So it's not "a drop in the push"; the drop itself is the push.

It's the electric field that is the push. More formally, Ohm's law is

$$ mathbf J = sigmamathbf E $$

That is, current density $mathbf J$ is proportional to the electric field $mathbf E$. The constant of proportionality $sigma$ is the material's conductivity.

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Detour: Why Ohm's law?

Remember that electric field is force per unit charge. So if there is an electric field within the material, the electrons, which are charged particles, feel a force and are accelerated in that direction.

This could lead you to think that in a constant electric field, the electrons are constantly accelerating, and thus the current is increasingly linearly in time. This is a more correct picture for electrons in free space, but not within a material. Within a material the electrons collide with other atoms and with each other (there's also thermal motion…), and so they actually move more like the beans in a Galton board. In this situation we also have gravity exerting a constant force on the beans, but due to hitting the obstacles, they move with an average constant velocity.

So, on average, the electric field causes the electrons to move with a constant velocity, and this means a constant current. So Ohm's law.

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Imagine your resistor is a cylinder, and we have a constant electric field $mathbf E$ pointing along the axis. Then by Ohm's law the current through the resistor is $I = iint mathbf Jcdot dmathbf A = iint sigmamathbf Ecdot dmathbf A = |mathbf E|sigma A$ where $A$ is the cross-sectional areal of the cylinder.

Voltage is defined as thintegral of the electric field, and so the voltage between the two terminals 2 and 1 of your resistor is

$$ V = int_1^2 mathbf Ecdot dboldsymbolell = |mathbf E|L = Ifrac L{sigma A} $$

where $L$ is the length of the cylinder. Or letting $R=L/(sigma A)$ be the resistance of our resistor,

$$ V = IR $$

which is Ohm's law in its familiar form for circuit theory.

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In short, if there is a difference in voltage between two terminals, then there is an electric field inside it. And the electric field that pushes the electrons and causes the current.

Only differences in voltage are meaningful. When we say that a given node in the circuit is at 24 V, we really mean that the integral of the electric field between that node and the ground node is 24 V. With no signal ground specified, the statement is meaningless. Also, if both terminals of your resistor are at 24 V, then there is no electric field between them, and so no current.

There is no need for a "push" after passing the resistor.

The voltage (the potential difference) across a resistor causes the electric "push" since this corresponds to a electric force fighting through the resisting material. As soon as the current is through, then there's no need for the push anymore. The charges can continue flowing along the wire after the resistor unrestricted since there is no more resistance, so no "push" is needed.

For an analogy, you can explain this with a water hose. If you squeeze the hose, you create a point of resistance. The pressure is now higher before and lower after this point - there is a pressure drop across this point. Naturally, this is only the case because the water before this point needs more "push" to pass the restriction than the water after.