Is the shell theorem only an approximation?
If you put a particle very close to the border, the force from matter very close to it will be very strong, as you say. But that is only a small portion of the shell; all the rest is pulling the other way, towards the center. The shell theorem guarantees that these forces cancel exactly.
Most of the mathematical formlities are dealt with on the wikipedia page you reference in your question. There, to prove the shell theorem, the shell is taken to have mass per unit area $\sigma$ and split it up into lots of coaxial rings, with the axis running along a diameter that goes through the test mass. $\theta$ is the angle between the diameter through the test mass and a line from the centre of the sphere to one of the rings.
Skipping to the main point, we can find the gravitational field generated by one of the thin rings at a test position that is a distance $r$ from the centre of a sphere of radius $R$ and mass $M$, where $s$ is the distance from the test position to a point on the ring is given by
$$dg = \frac{GM}{4Rr^2}\left( 1 - \frac{R^2-r^2}{s^2}\right)\ ds$$ which integrates to $$g = \left[\frac{GM}{4Rr^2}\left(s + \frac{R^2 - r^2}{s}\right)\right]^{s_{\rm max}}_{s_{\rm min}}.$$
For the full spherical shell the limits are $s_{\rm min} = R-r$ (when $\theta=0$) to $s_{\rm max}=R+r$ (when $\theta=\pi$ and the result is zero - this is the shell theorem and should work for any value of $r\leq R$.
However to answer your specific question, why doesn't the gravitational field due to the piece of the shell closest to the test point blow-up towards infinity as the test point gets very close to the surface of the shell and overwhelm the opposite (but clearly finite) field generated by the mass distributed over the rest of the sphere?
Look at the equation above and how it behaves when $r$ is very close to (but smaller than) $R$. In this case $s_{\rm min} \simeq 0$, and $(R^2-r^2)/s \simeq 2R$ resulting in a finite lower limit.
In words what is happening is that the amount of mass that is "infinitesimally close" to the test mass becomes infinitesimally small, ensuring that the gravitational effect of this mass does not blow up to infinity.
The shell theorem assumes a continuous distribution of matter in the shell.
If you came infinitesimally close to a real, physical shell you would discover that it, too, is made of particles. As you passed through the shell one of two things would happen:
You could crash into one of the particles and experience a non-gravitational force.
You could pass through a hole in the shell to the outside.
The second case is interesting. For a continuous shell there is a discontinuity between the force inside (zero) and the force outside (equivalent to a point mass at the shell's center). For a particle passing through a small hole in a spherical shell that discontinuity gets smoothed out. If the particle doesn't pass through the center of the hole, the smoothed-out gravitational force will include some component parallel to the surface of the shell; the details depend on exactly how "clumpy" the shell is.