The complete stress tensor, while accurate, is largely unnecessary for solving this problem, as it is a thin walled pressure vessel
Assuming the balloon is spherical, the strain can just be calculated from the current and initial radii.
$$\epsilon=\frac{r}{r_0}-1$$
The stress can be found using the modulus of elasticity:
$$\sigma=E\,\epsilon$$
The thin wall pressure equation can get you to pressure, if you know the thickness, by balancing outward pressure inside with the inward tension along a great circle of the sphere:
$$\pi\,r^2\,P=2\,\pi\,r\,\sigma\,t$$
$$P=\frac{2\,\sigma\,t}r$$
Because balloons get thinner as they stretch, the thickness will actually vary. Rubber typically has a poisson's ratio of 0.5 meaning it keeps a constant volume while being deformed. We can then calculate the thickness in terms of the radius:
$$t\,r^2=t_0\,{r_0}^2$$
$$t=t_0\,\left(\frac{r_0}{r}\right)^2$$
Putting them all together:
$$P=\frac{2\,E\,\left(r-r_0\right)\,t_0\,r_0}{r^3}$$
To see what this looks like, we can make a generic plot:
As you can see, there is a maximum pressure after which it becomes easier and easier to inflate the balloon. We can solve for this maximum pressure by equating the derivative with zero, solving for r, and plugging back in:
$$0=\frac{dP}{dr}=2\,E\,t_0\,r_0\left(\frac1{r^3}-3\frac{r-r_0}{r^4}\right)$$
$$r=\frac32\,r_0$$
$$P_{max}=\frac{8\,E\,t_0}{27\,r_0}$$
Of course this assumes a constant modulus of elasticity, which never holds true for a large enough deformation.
The difference between the "mouth hole" and the "needle hole" is that the mouth hole is reinforced and under low tension, while the needle hole is not reinforced and under high tension.
The mouth piece of a balloon, some might call it the "neck" of the balloon, generally includes a rolled section of material. This section is what I call it being "reinforced," since it is stronger than the rest of the balloon. Additionally, the neck is under low tension compared to the stretched fabric of the rest of the balloon.
So, why should any of that matter?
The tension on the surface of the balloon determines the response the balloon material will provide under a given load. When you poke a hole in, or open, the neck, not very much happens, because there is low tension (low force). However, when you poke a hole in the body of the balloon, the much higher tension there provides a larger response.
You can make an analogy to a stretched rubber band. If you stretch it a small amount, it is easy to perturb the rubber band a great distance. However, if the rubber band is stretched to near breaking, then even a small perturbation results in a snapping of the material, and the subsequent release of energy (both in motion of the pieces and in generation of sound).
Best Answer
My guess is that, leaving aside the tying off knot, the latex composing the ballon is not spread evenly. Balloons are made by dipping a spherical shape/mold into a vat of liquid latex and then lifting it out and allowing it to dry, so latex will flow downwards.
Also, they might increase the thickness of the balloon at the neck, to take the strain of tying the knot.
Latex balloons being manufactured.
As for equations of shape, I honestly have no idea of how these could be derived from first principles, sorry.