It's convenient to consider the variant of the 2D walk that has you take both
a horizontal and a vertical step at the same time, thus executing an ordinary
random walk on a larger lattice at a 45 degree angle to the first (the black
squares on a checkerboard). This makes your $x$ and $y$ coordinates independent
random variables, which they are not for the ordinary 2D walk.
Now the probability that $x=0$ at time $2n$ is ${{2n} \choose n}/2^{2 n}$ which is
approximately $const/\sqrt{n}$.
An "intuitive" plausibility argument for this is that the variance of $x$ increases by 1 at each step, so the standard deviation
is $\sqrt{n}$, i.e. there should be something like $\sqrt{n}$
possible values that have comparable probabilities, so
about $1/\sqrt{n}$ for each.
Thus the probability that both $x=0$ and $y=0$ at time $2n$ is the square of this
probability, which is approximately $const/n$.
Now the expected number of returns to the origin is the sum of these
probabilities, which is $\infty$ since the harmonic series diverges.
But if the probability of ever returning to the origin was $p < 1$, the
expected number of returns would be the finite number $p/(1-p)$ (think
of a sequence of independent coin-flips with probability $p$ of heads: the
expected number of consecutive heads is $p/(1-p)$).
In 3D, the analogous calculation (actually for a walk on a body-centred cubic
lattice) shows that the probability of return at time $2n$ is approximately
$const/n^{3/2}$. But this time the series is convergent, so the expected
number of returns is finite, and therefore the probability of ever returning
must be less than $1$.