I haven't seen this definition of the Mercator projection before but perhaps it makes sense. Let us assume that physically what is meant by the bladder picture is that once the bladder makes contact with the outside cylinder, the portion of surface which touches the cylinder remains fixed; it doesn't slide around. So the shape it had at the moment before contact is preserved.
Then we can see that the remaining portion of the bladder, the free portion, must take on the shape of a spherical bubble, because the pressure inside the bladder is only balanced when the skin is uniformly stretched, as only a spherical shape can do. Therefore the free portion of the bladder, being uniformly stretched, must accurately preserve the shapes of all physical features which are drawn on its surface.
Since the outside cylinder captures the local shape of the bladder at the moment it makes contact, the resulting projection preserves shapes in just the way that Mercator is designed for. Incidentally, the constraining cylinder need not be tangent to the initial sphere. Of course this is obvious because the bladder grows as a sphere until it makes first contact.
EDIT: I really had no idea what I was getting into with this problem. I think it's a very good math physics problem and unless I'm mistaken it's much harder than I would have thought. So I'm going to make a simplification or two, a few observations and a couple of guesses.
First, the physics becomes a little easier to visualize if we start off by considering a region rather near the North Pole. That way we're starting with a flat sheet. Then, we can replace the long pipe with a simple ring. The bladder is inflated until it makes contact with the ring. At that moment the portion of the ladder within the ring, initially almost a flat sheet, can be considered to be fixed on its circumference, and it will grow outwards as a bubble. At some moment in time the walls of the bubble become vertical. That is the moment when it would become fixed to the wall of the pipe (if the pipe were present). The problem now becomes simply to show that in the vicinity of the ring, at the moment the bladder becomes vertical, that the local deformation of the bladder is distortionless.
This simplification changes the problem less than one would think. If it is true for an arbitrary ring near the north pole, it is true for all rings, in particular a series of rings stacked one upon another. So it is true for the pipe - at least in the northern latitudes. We will put aside for now the question as to whether the physics leads to the same results in the mid to lower latitudes. (Althought it is almost trivially guaranteed to be true near the equator.)
Back to the ring. We consider the bladder to be inflated until it touches the ring, and at such a time it is large enough compared to the ring that the constrained portion is essentially a flat sheet. It should be fairly clear that the stretching is distortion-free up to this point. We wish to show that as the bubble deforms into the ring, it remains distortion-free near the ring until the moment when the walls of the bubble become vertical. Since the sheet is fixed laterally at the ring, this forces the seemingly strange result that the longitudinal stretching near the ring is zero - that once it makes contact, it simply tilts upwards without further stretching!
Yet the sheet must stretch somewhere: can it bee that the stretching is a maximum near the middle but goes to zero at the periphery?
This is in contradiction to what I claimed in my initial answer as to how only a spherical shape can be in equilibrium with a constant pressure. It seems I was wrong.
I am now going to propose that in this case, the physics can be satisfied if the curvature is locally spherical everywhere, but may change continuously. In that case, in order for the pressure to be everywhere balanced, the product of curvature and tension must be a constant. I have drawn her a series of nested spheres which I believe approximates the true form of the bladder expanding into a ring:
You can see that the curvature decreases near the dome, which means that the tension (and stretching) of the bladder is at its greatest near the North Pole. Near the point of constraint. This makes some sense if we imagine that the bladder is trying to accomodate the maximum volume for the least amount of stretching energy. In two dimensions the solution is of course contant stretching everywhere, leading to a circular profile. In three dimensions, since the area increases towards the outside, it is more economical for the bladder to stretch a little more in the middle. (The resulting curve looks something like a cycloid, but I do not believe it is.)
In any case, the interesting point is that if this model is correct, the stretching is everywhere distortionless. So at the point the bladder lays itself onto the inside of the pipe, shapes are preserved in their original form.
There is one sore point remaining: I cannot convince myself that the stretching must be everywhere distortionless...that the bladder is everywhere locally spherical. For a small square drawn anywhere on the bladder, there are two curvatures and two tensions: longitudinal and latitudinal. It is not clear to me why these two must be equal. The resistance to pressure is given by the product of tension and curvature, and I can imagine that the pressure might be balanced by the average of longitudinal and latitudinal resistance, without these quantities necessarily being equal.