Well, the wave function of the electron in the ground state of a hydrogen atom (and very similarly in other atoms) behaves like
$$ R(r) \sim \exp(-r / a) $$
where $a$ is the Bohr radius, effectively the radius of the atom. The exponential is in principle nonzero for an arbitrarily large $r$, so the electron may be found arbitrarily far from the nucleus at a nonzero probability.
Of course that the statement on Wikipedia is in principle correct. I don't believe that there is anything incomprehensible or ambiguous about that statement.
In practice, if we are already 100 Bohr radii from the nucleus which is much less than a micron, the probability already drops $\exp(200)$ times (because the wave function has to be squared). That's $P\sim 10^{-87}$, smaller than the inverse number of particles in the visible Universe, so you can be pretty sure that the electron (in a low-energy bound state of the atom) won't ever be found more than a micron from the nucleus. In most cases, it's not further than 3 Bohr radii from the nucleus.
But again, in principle, it may be anywhere. The exponentially small probability is similar to the probabilities that we tunnel through a classically impenetrable barrier. The probability is nonzero but negligible for thick enough barriers.
When talking about the stolen electron, one must realize that particles are constantly being stolen from and added to our bodies and two electrons can't even be distinguished from each other, so there is no reliable way allowed by the laws of physics that would allow one to say that "this electron is mine", "this electron is yours". Generic elementary particles, especially electrons, are being stolen all the time. It's particularly true if two conductors (pieces of metal) are in contact. In that case, the electrons immediatley become "shared". For two equally large pieces, within a very short moment of time, each electron has the same probability to be stolen as to be the from the original piece. This question really doesn't make sense because there's no invariant way to trace them.
You might try to "block" electrons that deviate too far from the nucleus. Absorb them if they cross a "red line", for example. You must realize that by adding the "absorber", however, you are also changing the rules of the game. The energy levels of the electron will be modified in the presence of both the nucleus and the absorber. One can't really separate these two things – the behavior of the electron inside the atom and its interactions with the absorber – because the absorber is an example of a measurement apparatus and those always have to influence the measured object, according to the most general principles of quantum mechanics.
The probability of finding the electron in some volume $V$ is given by:
$$ P = \int_V \psi^*\psi\,dV \tag{1} $$
That is we construct the function called the probability density:
$$ F(\mathbf x, t) = \psi^*\psi $$
and integrate it over our volume $V$, where as the notation suggests the probability density is generally a function of position and sometimes also of time.
There are two ways the probability $P$ can turn out to be zero:
$F(\mathbf x, t)$ is zero everywhere in the volume $V$ - note that we can't get positive-negative cancellation as $F$ is a square and is everywhere $\ge 0$.
we take the volume $V$ to zero i.e. as for the probability of finding the particle at a point
Now back to your question.
The node is a point or a surface (depending on the type of node) so the volume of the region where $\psi = 0$ is zero. That means in our equation (1) we need to put $V=0$ and we get $P=0$ so the probability of finding the electron at the node is zero. But (and I suspect this is the point of your question) this is a trivial result because if $V=0$ we always end up with $P=0$ and there isn't any special physical significance to our result.
Suppose instead we take some small but non-zero volume $V$ centred around a node. Somewhere in our volume the probability density function will inevitably be non-zero because it's only zero at a point or nodal plane, and that means when we integrate we will always get a non-zero result. So the probability of finding the electron near a node is always greater than zero even if we take near to mean a tiny, tiny distance.
So the statement the probability of finding the electron at a node is zero is either vacuous or false depending on whether you interpret it to mean precisely at a node or approximately at a node.
But I suspect most physicists would regard this as a somewhat silly discussion because we would generally mean that the probability of finding the elecron at a node or nodal surface is nebligably small compared to the probability of finding it elsewhere in the atom.
Best Answer
You are quite correct that atoms don't have a precise size. When defining the size of atoms we tend to use either bond lengths, if the atoms are reactive, or interatomic potentials for atoms that aren't reactive.
For example take Argon atoms, which are unreactive. The force between two argon atoms is well described by the London dispersion force, which in the case of Argon looks like:
Typically we get a minimum in the energy (at about 380pm in this case) then a hard core repulsion i.e. the energy rises steeply as you push the atoms past the minimum. We can take this as a measure of the size of the atom.
Where atoms react we can use the spacing in the molecules as a guide. For example the O to O distance in the O$_2$ molecule is 121pm giving us a radius for the Oxygen atom of a bit over 60pm.
However we will get different values for the atomic radii depending on how exactly we define it. For example Carbon forms single, double and triple C-C bonds, and they all have different lengths giving us different radii for the Carbon atom. This means the figures for the sizes of atoms are a guide rather than a precise value.
Wikipedia has a list of atomic sizes here that shows the different values for the radii obtained using different measures.