Can all canonical transformation be obtained through generation function approaches? The question can be formulated as following:
Suppose 
$$\delta \int_{t_1}^{t_2}{[p\cdot \dot{q} - H(p,q,t) ]dt} = 0$$
$$\delta \int_{t_1}^{t_2}{[P\cdot \dot{Q} - K(P,Q,t) ]dt} = 0$$
in which $$P = P(p,q,t), Q = Q(p,q,t)$$ is an invertible transformation.
Can we prove that there must exist a $\lambda$ and function $G(p,q,t)$ (or $G(p,Q,t)$, $G(P,Q,t)$, $G(P,q,t)$), such that 
$$\lambda[p\cdot \dot{q} - H(p,q,t) ] = [P\cdot \dot{Q} - K(P,Q,t) ] + \frac{dG}{dt}~?$$
 A: Before we begin, note first of all, that there exist various definitions of a canonical transformation (CT) in the literature, cf. e.g. this Phys.SE post. For instance, OP's last equation (v1) is called an extended canonical transformation (ECT) in Ref. 1.
OP is essentially asking (v1):

If we have a transformation
  $$\tag{A} (q,p)~\longrightarrow~ (Q,P)$$
  (with possible explicit time dependence) that transforms the Hamilton's eqs. into Kamilton's eqs., will it be an ECT, at least locally?

The answer is: No, not necessarily. For instance, the example in this Phys.SE post is a counterexample. This can be shown by a slight modification of the proof in eqs. (4)-(7) of my answer in order to allow an arbitrary scale factor $\lambda$.
For completeness, let us mention that the opposite of (A) is true: An ECT transforms the Hamilton's eqs. into Kamilton's eqs. This follows because their two action functionals $S_K=\lambda S_H$ are proportional to each other (up to boundary terms). Therefore the EL eqs. (=Hamilton's eqs.) for $S_H$ corresponds to the EL eqs. (=Kamilton's eqs.) for $S_K$.
References:


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*H. Goldstein, Classical Mechanics, Chapter 9. See text under eq. (9.11).

