What's the reason behind the current remaining the same after passing by a resistance? I've been wondering why does this really happen, I mean by intuition if electrons are driven by EMF (ignoring wire's resistance), $n$ coulombs would pass by a point per second, until they encounter something that slows them down thus the rate of flow would change. Why does current remain the same?
One answer I saw somewhere that made sense to me is that it indeed slows electrons down, but electrons lose some of their energy to compensate the loss of velocity in a way that would bring the current back to the constant current, and this lose of energy is called drop in voltage and that's why voltage decreases when running over some resistance, is this true?
 A: Current us a measure of how much charge is passing a given point (or cross section) of a wire.
If the currents were not equal at all points in a simple circuit,  there would have to be charges entering or exiting the circuit. This however does not happen.
Water pipe analogy: current is something like liters per minute that pass through a certain point. If there are no leaks or additional pipes joining, at each point there has to be the same water flow in liters per minute. 
A: Compare the conductor with a water pipe: the flow rate in the water pipe is the same everywhere (assuming there are no branches or leaks or whatever). Now introduce a constriction in the pipe, or a valve that's half closed. In this new situation the flow rate in the pipe is lower, but it is lower everywhere in the pipe: both before and after the constriction. The flow rate is still the same everywhere in the pipe (but not the same as before).
The same is true in a conductor: in a conductor with a resistance the current (flow rate) will be lower than in a conductor without (given the same voltage), but each conductor will have the same current everywhere.
A: 
What's the reason behind the current remaining the same after passing
  by a resistance?

Because, due to the redistribution of charges (and voltages), the electric field inside a resistor is stronger than the electric field inside a wire.
This redistribution happens automatically. Initially, the electric field is evenly distributed and the electrons in the wire move faster than the electrons inside the resistor. As a result, electrons accumulate in front of the resistor and positive ions accumulate behind it. This increases the voltage and the electric field across the resistor, causing the electrons inside the resistor to move faster. 
Since the voltage of the battery stays the same, the increase of the voltage across the resistor leads to the decrease of the voltage and, therefore, the decrease of the electric field in the wire, causing the electrons in the wire to move slower.
As the electrons in the resistor move faster and faster and the electrons in the wire move slower and slower, at some point, their speeds will equalize. At this point, the redistribution of charges will stop and the current in all parts of the circuit will be the same. 
A: Yes, it is true. 
Charge cannot disappear. Or appear. This is Kirchhoff's current law. For a steady current, all charge that enters any point each second must also leave that point each second:
$$\sum i_{in}=\sum i_{out}$$
Otherwise charge would accumulate at that point. And eventually the enormous total charge that is accumulated will be large enough to repel any further entering charge - large enough to balance out the battery voltage - and all current would stop flowing. Since this doesn't happen, charge isn't accumulating anywhere, so Kirchhoff's current law must be true.
So what is happening?
When turning on your battery / voltage source, charges are being "pushed" forward by the battery voltage. The first charge hurries with almost the speed of light. Very, very quickly it reaches the resistor. Here it is slowed down. All the charges behind it now have to queue up and wait in line - they slow down as well. Soon they all move at the exact same speed.
When exiting the resistor the charge continues with whatever speed it came out with - which naturally is the same as the speed of all those waiting in line.
So, now, suddenly, all charge moves at the same speed. In other words, the current (amount of charge passing per second) is the same everywhere. 
The push on the charges, on the other hand, is large before the resistor and zero (relatively) after the resistor. In the same way that the pressure in a water hose is large, but as soon as the water is out, there is no pressure any more. The pressure is released. Likewise, the voltage is dropped.
The supplied voltage is "spent" across the resistor, so to speak, and we talk about it as a voltage drop. And the full "pressure" / voltage supplied by the battery will be spent. In other words, the entire supplied voltage must be dropped and spread out over all components along the way in the circuit. Which leads to Kirchhoff's other law, his voltage law or loop law: all "spent" voltage in any circuit loop must necessarily equal what is being supplied somewhere else: $$\sum v_{added}=\sum v_{dropped}$$
A: I have found a nice explanation called the Drude Model, a simple mechanical model proposed by the German physicist Paul Drude in 1900. It is based on the idea that the charge carriers bounce around randomly at high speeds, with an average momentum of
$$\langle \vec{p} \rangle = q\vec{E}\tau$$
where $q$ is the charge of a carrier (i.e. the electron charge), $\vec{E}$ is the electric field vector and $\tau$ is the average time between bounces.
Quoting from the wikipedia article, use the following substitutions
$$\langle \vec{p} \rangle = m\langle \vec{v} \rangle,\; \vec{J} = nq\langle\vec{v}\rangle$$
where $m$ is the electron mass, $n$ is the number density of the charge carriers, $\langle \vec{v} \rangle$ is the average velocity, and $\vec{J}$ is the current density. This gives us:
$$\vec{J} = \left(\frac{nq^2\tau}{m}\right)\vec{E}$$
Now consider a resistor with the oriented cross-section area $\vec{S}$. Multiply the above formula together with $\vec{S}$ to get the current through $\vec{S}$:
$$I = \vec{J}\cdot\vec{S} = \left(\frac{nq^2\tau}{m}\right)\vec{E}\cdot\vec{S}$$
For simplicity, assume $\vec{E}$ and $\vec{S}$ are constant and parallel, and that $|\vec{E}|=\Delta U/L$, where $\Delta U$ is the voltage difference and $L$ is the length of the resistor. This gives us the current in terms of the voltage difference times a factor:
$$I = \left(\frac{nq^2\tau}{m}\right)\frac{\Delta U\,S}{L} = \left(\frac{nq^2\tau S}{mL}\right)\Delta U$$
This is Ohm's law if we identify $\left(\dfrac{nq^2\tau S}{mL}\right)=\dfrac{1}{R}$.
Note: the above only shows the steady-state situation. It doesn't really explain how the local voltages and Kirchoff conditions reach this steady state after the voltage is applied initially. But the Drude model can also handle time-varying dynamics. Please see the Wikipedia article for details.
