Why do $x - ct = 0$ and $x' - ct' = 0$ imply that $(x - ct) = \lambda (x' - ct')$ in special relativity? From the book, Relativity: The Special and General Theory (PDF) by Albert Einstein - Appendix 1:

A light-signal, which is proceeding along the positive axis of $x$, is  transmitted according to the equation $$ x = ct $$ or $$ x - ct = 0. \hspace{2em}(1)$$ Since the same light-signal has to be transmitted relative to $K'$ with the velocity $c$, the propagation relative to the system $K'$ will be represented by the analogous formula $$ x' - ct' = 0. \hspace{2em}(2)$$ Those space-time points (events) which satisfy $\text{(1)}$ must also satisfy $\text{(2)}$. Obviously this will be the case when the relation $$ (x' - ct') = \lambda (x - ct) \hspace{2em} (3) $$ is fulfiled in general, where $ \lambda $ indicates a constant; for, according to $\text{(3)}$, the disappearance of $(x - ct)$ involves the disappearance of $(x' - ct')$.

Can someone explain why we must conclude the equation (3) from equations (1) and (2)? I am unable to see why we need to introduce the constant $\lambda$.
If $ x - ct = 0 $ and $ x' - ct' = 0 $, then I get $ x - ct = x' - ct' = 0 $. I see no need for a constant $ \lambda $. Can someone explain to me why must I add a constant $ \lambda $ in equation (3).
 A: Eq. (3) is a consequence of the symmetry imposed by the principle of relativity.
I hope my proof is what you are looking for.
First of all, we assume $(x'-ct')$ can be expressed as a function of 
$(x- ct)$,
$$x' - ct' = \Phi(x-ct) \quad ,$$
with $\Phi$ being at least invertible and $C^1$.
Side remark : From the technical point of view this follows from the fact that space-time is a differentiable topological manifold with a smooth atlas, however we don't need so much rigour here.
The systems $K$ and $K'$ are equivalent in all respects. 
Therefore the inverse function must have the same functional form up to a constant
$$x - ct = \Phi^{-1}(x' - ct') = \alpha \, \Phi(x'- ct')
\qquad \quad (*)$$
Note 1 : Equivalence doesn't mean triviality though.
We must have $\Phi(0)=0$
(the speed of light is the same for all inertial observers), 
and the functional form is the same "up to a constant" because a change in the units of measure doesn't spoil the physical equivalence.
Note 2 : $\alpha=0$ has no physical meaning. 
If it was true, 
$K$ would see any particle moving at the speed of light, 
even if for $K'$ the latter was slower.
Eq. $(*)$ is true only if $\Phi$ is a linear function.
Let's see why.
Consider a value $a$ and map it to $b=\Phi(a)$. 
As discussed above one has:
$$a = \Phi^{-1}(b) = \alpha \, \Phi(b) = \alpha (\Phi \circ \Phi) \, (a)$$
From the arbitrariness of $a$ we conclude that $\alpha (\Phi \circ \Phi)$ is the identity operator. Hence we have:
$$\Phi\left(\frac{a}{\alpha}\right) = 
\Phi \circ \Phi \, (b) = 
\frac{1}{\alpha} b = 
\frac{1}{\alpha} \Phi(a) 
$$
Calling $\beta = 1/\alpha$:
$$\Phi(\beta \, a) = \beta \, \Phi(a)$$
In other words $\Phi$ is a homogeneous function of degree $1$.
From the Euler's theorem of homogeneous functions:
$$\Phi(a) = a \, \frac{d}{da} \Phi (a) $$
Deriving with respect to $a$ we get:
$$ a \frac{d^2}{da^2} \Phi (a) = 0 
\quad \rightarrow \quad
\Phi(a) = \lambda a  + \nu
\quad \text{for } a \neq 0
$$
However the invariance of the speed of light sets $\Phi(0)=0$, 
so we get the solution (now valid $\forall \, a$):
$$\Phi(a) = \lambda \, a$$
Back to our problem,
$$x' - ct' = \lambda (x - ct) \quad .$$
A: Equations $(1)$ and $(2)$ specify two trajectories in frames $K$ and $K'$ respectively. Since they are trajectories, $x$ and $t$ here are obviously not independent, and their relationship is specified by eq. $(1)$. Similarly $x'$ and $t'$ are related by eq. $(2)$. So these two equations specify two subsets of spacetime, the first one seen through $K$ and the second one through $K'$. Constancy of the speed of light requires these two sets to be the same. 
Our goal, on the other hand, is finding a transformation $(x, t) \rightarrow (x', t')$ that transforms any couple $(x, t)$ in $K$ into a couple $(x', t')$ in $K'$. Any couple, not only those satisfying $(1)$. The requirement of constancy of the speed of light imposes that when this transformation is applied to eq. $(1)$, it has to become eq. $(2)$.
Einstein here is simply saying that if we impose the much stronger requirement that eq. $(3)$ is satisfied by every couple $(x, t)$ with its transformed $(x', t')$, then eq. $(1)$ gets obviously transformed into eq. $(2)$.
I want to stress here that $(3)$ is not a tautological consequence of $(1)$ and $(2)$ with an added superfluous constant: indeed, $(1)$ and $(2)$ only specify two sets of spacetime points in frames $K$ and $K'$ respectively, while $(3)$ is a relation that Einstein is imposing on every spacetime point, also those not belonging to sets $(1)$ and $(2)$ (notice that, after eq. $(3)$, he writes that it is imposed to be "fulfilled in general"). 
Now you could ask: But why should we use specifically requirement $(3)$, and not one of the infinitely many others that would equally ensure that $(1)$ gets transformed into $(2)$? Well, see @momi94's answer for that. 
A: All the book is saying (at this stage) is that this is obviously a solution.  There is no suggestion at this stage that it is the only solution.
Equally obviously $(x-ct) = λ(x'-ct')^n$ is also a solution for any non-zero value of λ and any non-zero value of n.  Many other solutions are also available.  
Any addition of two or more different left or right hand sides is also a solution, and any multiplication of two or more solutions is also a solution.
As advised above, I too suggest you read on, remembering the limit of the logic.
