Deriving the equations for a moving inertial reference frame I assume $c=1$ in the following derivation:
In order to derive the equations for a moving inertial reference frame, I immediately wrote down the following:
$$ x'=Ax+Bt, \tag{1}$$
$$t'= Dx+Et. \tag{2}$$
In order to solve it I would need 4 independent equations. Here they are:


*

*Since the speed of light is constant in all reference frames, it follows that if $x = t$, also $x' = t'$, therefore


$$At+Bt=(Dt+Et) \overset{(x=t)}{\implies} A+B=D+E. \tag{3}$$


*I can reverse the reference frame and the physics should be the same hence 
$$x=0 \implies x'=-vt' \implies B=-Ev. \tag{4}$$

*Finding the $x'$ component of the point $A(\frac{1}{1-v},\frac{v}{1-v})$ gives:
$$Av+B= -Av. \tag{5}$$

*Finally finding the $t'$ component of $A$ gives (I'll do this one explicitly):
$$t'= \frac{D+Ev}{1-v}. \tag{6}$$
From the diagram one can read off using the Pythagorean theorem that:
$$t'= \sqrt{\left( \frac{1}{1-v} \right)^2 +\left( \frac{v}{1-v} \right)^2 } = \frac{\sqrt{1+v^2}}{1-v}$$
  $$\implies D+Ev= \sqrt{1+v^2}. \tag{7}$$

From these equations one easily arrives at the desired result ie
$$x'=\frac{x-vt}{\sqrt{1-v^2}} \; \text{and} \; t'=\frac{t-vx}{\sqrt{1-v^2}} \tag{8}$$
All this seems to be correct. However considering the equation $(6)$ and putting back the $c$'s in it one arrives at the equation
$$ D+Ev= \sqrt{1+v^2} \quad (!) \tag{9}$$ 
First of all this dimensionally doesn't make sense. Secondly if you calculate and find the coefficients you don't get the correct answer. Intuitively I know that this equation has to be $D+Ev= \sqrt{1+v^2/c^2}$ so that everything works perfectly but I don't know why this has to be so and I cannot show it by reasoning physically. I fell in my guts that there is something fishy about using Pythagorean theorem but I don't know what went wrong exactly. If I just say that the use of Pythagorean is wrong, then I cannot explain why it gives the correct answer when using $c=1$. Such a coincidence seems to be highly unlikely.
Edit: I've made a major typo in the diagram you should swap $x=0$ with $t=0$ and $x'=0$ with $t'=0$!
 A: You work too hard and the idea of setting $c=1$ may make problems.
I copy here the equations that you obtained and that I found correct. So,
1. The speed of light is the same in each frame implies
$$Act + Bt = c(Dct + Et), \overset{(x=t)}{\implies} Ac + B = Dc^2 + Ec. \tag{i}$$


*Reversing the frames gives indeed


$$B = -Ev. \tag{ii}$$


*Also in the frame $(x,t)$ the origin $x'=0$ of the frame $(x',t')$ moves at velocity $v$


$$B = -Av. \tag {iii}$$
Notice also that from $\text {(ii)}$ and $\text {(iii)}$ one infers
$$E = A, \tag{iv}$$
and introducing all these relations in $\text {(i)}$,
$$Ac - Av = Dc^2 + Ac \overset{(x=t)}{\implies} D = -\frac {Av}{c^2} \tag{v}$$.


*Now putting all these things together I rewrite your transformations $(1)$ and $(2)$


$$x' = A(x - vt), \ \ \ ct' = A(-\frac {vx}{c} + ct). \tag{vi}$$
Whatever remains is to find A. In this task, the interval conservation is bound to help,
$$c^2t'^2 - x'^2 = c^2t^2 - x^2. \tag {vii}$$
So let's do the calculus,
$$A^2 \left[(-\frac {vx}{c} + ct)^2 - (x - vt)^2 \right] = c^2t^2 - x^2. $$
Doing the calculus you get
$$A = \frac {1}{\sqrt {1 - \frac {v^2}{c^2}}} \tag{viii}$$
So let's now rewrite the transformations
$$x' = \frac {x - vt}{\sqrt {1 - \frac {v^2}{c^2}}}, \ \ \ ct' = \frac {ct -\frac {vx}{c}}{\sqrt {1 - \frac {v^2}{c^2}}}. \tag{ix}$$
Now, I saw that you have a problem with the calculus of your expression $(6)$ but you didn't say what it means. Anyway, for $D$ see my formula $\text {(v)}$ and for $E$ my formula $\text {(iv)}$. You can rely on them, they are obtained in a simple way.
A: I am sorry, but I have no time to find your mistake for you. However, I can give you a tip that I am 100% sure would work.
Your mistake can be very easily found by dimensional analysis. Start from the beginning of your derivation and check all formulas. The first one which makes no sense due to dimensions is wrong.
You can just say that because $c$ is one in your calculations, you define $v$ do be what is actually $v/c$. That would give correct results by simple substitution $v \rightarrow v / c$, but I am under impression that is not good enough for you, though I can't understand why.
