Why do quantum effects of gravity become important at the Planck scale? The standard heuristic argument for why quantum effects of gravity become important at the Planck scale is to consider the length scales at which both quantum field theory (QFT) and general relativity (GR) become crucial in order to explain physical phenomena. 
For QFT this is when the length scale is of order the Compton wavelength of a particle, $$l_{c}=\frac{h}{mc},$$ since if one attempts to confine a particle within this length then it is possible for pair creation to occur and so the concept of a particle breaks down and QFT is required. 
For GR, this is when the length scale is of order the Schwarzschild radius of a particle, $$l_{s}=\frac{2Gm}{c^{2}},$$ since compressing the mass of a particle to within this radius results in the formation of a black hole which requires GR to understand its behaviour.
As such, one expects that when these to lengths are of the same order, i.e. $l_{c}\sim l_{s}$, that the quantum effects of gravity become important. This occurs when $$\frac{h}{mc}\sim\frac{2Gm}{c^{2}}\Rightarrow m^{2}\sim\frac{hc}{2G}\sim m_{P}^{2}$$ That is when the mass of the particle is of the same order as the Planck mass.
What I'm unsure about, is why the Compton wavelength? Why not the de Broglie wavelength as this is the length scale at which the quantum nature of an object becomes evident? 
Is it simply because QFT is consistent with special relativity and standard quantum mechanics is not and so it is the scale at which QFT becomes crucial that sets the scale for quantum gravity?
 A: From the QFT of prof.Tong 

We learn that particle-anti-particle pairs
  are expected to be important when a particle of mass
  m is localized within a distance of order
λ=h/mc
At distances shorter than this, there is a high probability that we will detect particle-anti-particle pairs swarming around the original particle that we put in.  The distance λ is called the Compton wavelength
  .  It is always smaller than the de Broglie wavelength

One has to remember that c is the maximum velocity due to special relativity and thus the smallest wavelength of a specific particle of mass m.
He continues:

If you like, the de Broglie wavelength is the distance at which the wavelike nature of particles becomes apparent; the Compton wavelength is the distance at which the concept of a single pointlike particle breaks down completely

Quantum field theory is based on relativistic quantum mechanics: creation and annihilation operators operate on free particle wavefunctions, of the Dirac field for fermions, the Klein Gordon for bosons and the quantized Maxwell equations for photons. These free particles are considered point particles, and the QFT expansions of the solutions depend on the fact that higher order corrections are of smaller value than the first order.  there is a high probability that we will detect particle-anti-particle pairs swarming around the original particle that we put in.  means that the series expansion breaks down and the QFT calculations dependent on point like particles is no longer reliable.
At the same time , classical general relativity calculations break down, ending in singularities, above  the Schwarzschild radius of a particle,

What I'm unsure about, is why the Compton wavelength? Why not the de Broglie wavelength as this is the length scale at which the quantum nature of an object becomes evident?

The maximum velocity is c for a particle of rest  mass m, and thus describes the maximum possible momentum  of a deBroglie wave 

Is it simply because QFT is consistent with special relativity and standard quantum mechanics is not and so it is the scale at which QFT becomes crucial that sets the scale for quantum gravity?

QFT is based on standard quantum mechanics. Its creation and annihilation operators act on relativistic wavefunctions of point particles in order to describe the Feynman diagrams used in the calculations. It is really the scale where QFT breaks down as a perturbative expansion, and the underlying basis of point particles . 
On the other hand it is expected that quantization of gravity will get rid of the singularities of classical general relativity. Already in the Big Bang model a fuzzy region is introduced where effective quantum mechanics is used  for the calculations. At Planck mass energies there are no longer the standard model point particles, and in various models a Planck mass is used as the particle  in expansions of quantized gravity, ( see note 96 in link) but the whole thing is not rigorous . A definitive quantization of gravity will define what sort of perturbative expansions will be valid at those energy scales.
A: User anna v has already given a correct answer. In this answer we try to summarize.
In a nutshell, the Planck scale of quantum gravity is determined by the 3 physical constants $G$, $c$ and $h$. 


*

*When the wavelength $\lambda$ becomes of the order of the Schwarzschild radius, the rest energy $mc^2$ becomes of the order of the gravitational energy $Gm^2/\lambda$. 

*When the wavelength $\lambda$ becomes of the order of the Compton wavelength, the rest energy $mc^2$ becomes comparable with the energy $hc/\lambda$ of a quantum. 
In contrast, the de Broglie wavelength $h/|{\bf p}|$ lacks information about relativity theory, and fail to identify the pertinent characteristic scale.   
