To understand this explanation, you need to understand Fourier decomposition of the electromagnetic field.
In any homogeneous medium, any electromagnetic field can be thought of as a linear superposition of plane waves, all in different directions. Because they run in different directions, the phase delays they undergo in propagating from, say, your aperture to another, parallel plane are all different. Therefore the wavefront gets "scrambled" owing to these direction-dependent phase delays. This interference between the different plane wave components of the electromagnetic field is what we commonly call "diffraction". I further explain this idea, as well as draw some diagrams in this answer here as well as this one here.
So with this introduction in mind, let's look at your paragraph. For simplicity, assume only one transverse direction and one axial (in the direction of propagation) direction. Let's also assume scalar optics, i.e. that the electromagnetic field is well represented by the behaviour of one of its Cartesian components, so that we can do Fourier optics on scalar field.
So we have a uniformly lit aperture of width $a$. Its transverse profile is therefore the function ${\rm rect}(2 x/a)$ where ${\rm rect}(x) = 1;\,|x| \leq 1$ and ${\rm rect}(x) = 0;\,|x| > 1$. We take a Fourier transform to find the superposition weights of each plane wave component, because each such component has a transverse variation $\exp(i\,k_x\,x)$ where $k_x$ is the Fourier transform variable with units of reciprocal length. The Fresnel distance is, as the paragraph says, simply the axial distance needed for this spread to double the beam width. So it is a rough measure of how quickly the light spreads.
So this is how the "divergence" arises from diffraction, i.e. the interference between an optical field's plane wave components as they propagate. Also $\sin\theta = k_x/k$ where $k = 2\pi/\lambda$ defines the angle that this plane wave component makes with the axial direction. We take the Fourier transform, we find that there is a spread of $k_x$ values such that the plane wave components most skewed to the axial direction make an angle without direction of roughly $\lambda/a$. So, owing to these skewed components, the field's energy spreads out.
The beam width diverges slowly at first and then, after an axial distance of several Fresnel distances, the divergence speeds up so that the propagation becomes well modelled by the cone of rays diverging from the centre of the aperture. Indeed if you plot contours of constant intensity, they are hyperbolas which begin at right angles to the aperture but bend so that their asymptotes are the cone defined by ray theory. The Fresnel distance defines how far the "knee" of the hyperbol is from the aperture.
For your question:
Is it not that the validity holds when all objects are comfortably larger, and not smaller, than the wavelength of light?
This is in general right, but it breaks down near focuses and in situations like this where we are near and aperture and if the aperture is comparable to the light wavelength. In this case you should be able to understand from the Fourier analysis the reciprocal relationship between the aperture width and the angular spread.
Best Answer
This is really just a geometry problem since we can not use Fraunhofer approximation in general. The solution depends on what you mean by "width" of the central maximum. If you take it to be the distance between the two adjacent minima (full width) then the solution is:
$$z_{\text{Fresnel}} = \frac{a^2}{4 \lambda} - \frac{\lambda}{4}$$
Alternatively the full width half maximum would probably have a factor of 2 in front or something like that. Note that for $\lambda \ll a$ the result becomes what LC7 wrote in his answer (who calculated full width half maximum), but it is not the most general form.
Here is my working. The crucial step is that at the minimum the interference is fully destructive, so for every point in the aperture exists another point in the aperture that is $\pi$ out of phase. Hence the phase difference at the minimum between edge of the aperture and the centre of the aperture is $\pi$ (see picture). The rest is then just doing the geometry.
EDIT in response to comments
I will try to give an explanation of how this fits with the question.
I calculated this distance above, fixing the ambiguous "width" to be the full width, as I explained (as opposed to full width half maximum). My derivation does not make any approximation other than scalar diffraction theory.
The OP talks about "beams" here. This is only valid in the Fraunhofer approximation, i.e. the far field limit. Why? Interference/diffraction causes a complicated wave field behind the aperture. Far away from it it looks a bit simpler, since only the linear phase terms will be significant. Now linear phase terms give you plane waves. These are the "beams" you are talking about. However the Fresnel distance does not lie in the Frauhofer regime region!! So unless you make further approximations this is not a valid approach.
Again this is a property of the diffraction pattern in the Fraunhofer limit and not true in general, which is where the whole misconception arises.
Comment about the comments
From what I gather from the comments people are just trying to find a derivation of some particular formula they learnt in some course. If that is the case I think the question should be closed. I think I have explained all conceptual aspects of the problem and am open to further suggestions about how to improve that, but I won't go into throwing around formulae without concepts.