How to derive $E=mc^{2}$?

A simple derivation that is accessible to lay people who can only do primary school level math, starts from the fact that a pulse of electromagnetic radiation with energy $\mathbf{E}$ has a momentum of $\dfrac{\mathbf{E}}{\mathit{c}}$. In addition, one assumes conservation of momentum. We do a thought experiment involving a closed box containing two objects of mass $\mathtt{M}$. No external forces act on the box, therefore the momentum of the box is conserved. This means that whatever happens inside the box, if the box was at rest initially, it will stay at rest. In particular, it is impossible for the center of mass of the box to move.

We then consider the following process happening inside the box. One object emits a pulse of light of energy $\mathbf{E}$ that is completely absorbed by the other object. Due to the recoil of the light, the masses will move, but the velocities will obviously be very small (for macroscopic objects). We can thus safely make the assumption that classical mechanics will be valid after the pulse of light has been absorbed. If you want to be very rigorous, you must take the limit of $\mathtt{M}$ to infinity so that classical mechanics will be exactly correct.

Due to conservation of momentum, in the original rest frame of the box, the center of mass of the box will not change. Now classical mechanics doesn't give a correct description of relativistic phenomena like light, but it can be used to describe the situation in the box both before the pulse of light was emitted and after it was absorbed.

What is clear is that the object that emits the pulse of light will move away from the other object with the velocity of $$v = \dfrac{\mathbf{E}}{\mathtt{M} \cdot \mathit{c}}$$. If the other object is a distance of $\mathit{L}$ away, then it will take a time of $T = \dfrac{\mathit{L}}{\mathit{c}}$ before the other object absorbs that pulse of light. So, it seems that the center of mass will shift by $\frac{1}{2} v\cdot T = \frac{1}{2} \dfrac{\mathbf{E} \cdot \mathit{L}}{\mathtt{M} \cdot {\mathit{c}}^2}$ because of the time lag between when the object emitting the puls of light starts to move and when the object absorbing the pulse of light starts to move.

But, of course, a closed box upon which no external forces act cannot move all by itself. The masses must have changed due to the transfer of energy. Since classical mechanics is assumed to be valid we also have conservation of mass, so the sum of the masses of the objects will not have changed. If the emitting object has lost a mass of $d\mathtt{M}$, the receiving object will have gained a mass of $d\mathtt{M}$. A transfer of a mass of $d\mathtt{M}$ from the emitting object to the receiving object will have shifted the center of mass by an amount of $\mathit{L} \dfrac{d\mathtt{M}}{2\mathtt{M}}$ in the opposite direction as the shift due to the motion. For the two effects to cancel each other out, requires that $$\mathit{L} \dfrac{d\mathtt{M}}{2\mathtt{M}} = \frac{1}{2} \dfrac{\mathbf{E} \cdot \mathit{L}}{\mathtt{M} \cdot {\mathit{c}}^2} \implies \mathbf{E} = d\mathtt{M} {\mathit{c}}^2$$.

So, transferring the energy via the pulse of light led to mass being transferred according to $\mathbf{E} = \mathtt{M} {\mathit{c}}^2$. Due to conservation of energy and the ability of energy to transform from one form to another (e.g. the pulse of light will become heat, chemical energy or whatever), one can then argue that any transfer of energy implies a transfer of the equivalent mass according to $\mathbf{E} = \mathtt{M} {\mathit{c}}^2$. This in turn then leads to the conclusion that the mass of an object is simply the total energy content of the object divided by ${\mathit{c}}^2$.

The answer really depends on how much you assume. You could start with the relativistic action, $$ S=-mc\int d\tau $$ where $c^2d\tau^2=dx^\mu dx_\mu$, such that the Lagrangian is $$ L=-mc\sqrt{1-(v/c)^2} $$ whence the momentum follows from $p=\partial L/\partial v$ and the total energy is $E=pv-L$--this leads directly to \begin{align} E&=\gamma mc^2=\left(1-v^2/c^2\right)^{-1/2}mc^2\tag{1}\\ p&=\gamma mv=\left(1-v^2/c^2\right)^{-1/2}mv\tag{2} \end{align} Where you can then use the first to solve for $v$ and insert into the latter to get the expected relation for $p=0$ (inserted after solving).

You could also start with (1) and assume $v\ll c$ such that you can use the approximation, $$ (1-x)^n\approx1-nx $$ such that $$ E\approx\left(1-\frac{v^2}{2c^2}\right)mc^2=mc^2+\frac12mv^2 $$ where the last term equals 0 for a stationary particle, leading again to the famous $E=mc^2$.

which I think is the unit for Newtons, not energy.

The SI unit of energy is the Joule, defined as $$ \rm Joule=1\,kg\,\frac{m^2}{s^2}=1\,N\,m $$ so $mc^2$ is indeed an energy, not force.

And why is there no constant of proportionality ? Did Einstein just set it up perfectly so that there wouldn't be one?

Einstein didn't "set it up" (which would imply that he made it up), he derived the equation. There is indeed a constant of proportionality, it is 1.

First a note not to forget that $E=m\,c^2$ is only a special case of the more general formula $\frac{E^2}{c^2} - p^2 = m^2 c^2$ for the Lorentz-invariant square "length" of the momentum four-vector in terms of the rest mass $m$.

That matter out of the way, there are two methods springing to mind:

Method 1: A neat way to do this is to imagine light in a massless box with perfectly reflecting mirrors in it and to analyse what happens when you shove the box.

Suppose this box is at first at rest with respect to your inertial frame. At equilibrium, the bouncing light exerts equal pressure on all mirrors.

However, now we wish to make the box move at some small, constant speed $v\ll c$ relative to our frame. As the box begins to move, the light reflecting off the hinder end of the box (in the sense defined by the motion) is blue shifted as it meets the mirror at the end of the box and that reflected from the forward end of the box is red shifted. So the pressures now differ, with the result that you need to supply impulse in the direction of motion to allow the light to reach a steady state moving at constant velocity relative to your frame. When you do the calculation, you find that this impulse is equal to $E\,v/c^2$. So you therefore ascribe a "resistance" to shove measured by an "inertia" $E/c^2$. I show how to do this calculation with photons (easy) here. One can also do a fully classical using the Lorentz transformation (much harder and messier) and the results are the same. Some day I must work this full analysis into a Physics SE answer, but have as yet not found occasion to do so: the technique is very like that in my answer here which describes "squashing light" in a cavity with moving mirrors.

Note that, although we are calculating, using classical, nonrelativistic mechanics, the force needed on the box, we are actually working out the limiting behavior as $v\to0$: we are considering a system moving at velocities as small as we please. So this method shows that we must ascribe a "mass" $E/c^2$ to a quantity of energy if special relativity and classical mechanics are to be mutually consistent, i.e. equal in the limit as $v\to0$. Thus we avoid problems associated with discussion of so called "relativistic" mass, which can yield misleading results, although it yields by chance the right answer in Jimmy 360's answer.

Method 2: This is the method initially used by Einstein in his first paper after his big special relativity paper of 1905:

A. Einstein, "Ist die Trägheit eines Körpers von seinem Energieinhalt abhängig?", Ann. der Phys. 18:639,1905 English translation "Does the Inertia of a Body Depend upon its Energy-Content?"

This is very like the method of Count Iblis's Answer