For any $n \ge 0$, let
$$f_n = {\bf Pr}\big[ S_0 =n \land \exists t > 0, S_t = 0 \big]$$
be the probability for a random walk start at location $n$ returns to origin in some finite time $t$.
Let $q = 1-p$ and $\displaystyle\;\mu = \frac{q}{p}\;$. It is easy to see $f_n$ satisfies a recurrence relation
$$f_n = \begin{cases}
1, & n = 0,\\
p f_{n+2} + q f_{n-1}, & n > 0
\end{cases}\tag{*1}$$
The corresponding characteristic equation
$$\begin{align}
p \lambda^2 + q\lambda^{-1} - 1 = 0
&\iff p(\lambda^2 - 1) - q(1-\lambda^{-1}) = 0\\
&\iff (\lambda^2 + \lambda - \mu)(\lambda - 1) = 0
\end{align}
\tag{*2}
$$
has three roots,
$$1,\quad -\frac{1+\sqrt{1+4\mu}}{2}\quad\text{ and }\quad\frac{\sqrt{1+4\mu}-1}{2}$$
This implies
$$f_n = A + B \left( -\frac{1+\sqrt{1+4\mu}}{2} \right)^n + C \left(\frac{\sqrt{1+4\mu}-1}{2}\right)^n$$
for some suitably chosen constants $A, B, C$.
Notice among the three roots, the $2^{nd}$ root is largest in magnitude and $< -1$. If $B \ne 0$, then $f_n$ will be negative for some sufficiently large $n$. This contradicts with the nature of $f_n$ being the probability of some event. This means
$B = 0$. What happens to $A$ and $C$ depends on $\mu$.
If $\mu < 2$, the average drift of the random walk $r = 2p - q = p(2-\mu) > 0$.
For $t$ not too small, we can approximate $S_t$ by a normal distribution with mean
$r t + n$ and standard derivation $\sigma = \sqrt{4p^2 + q - r^2} = 3\sqrt{pq}$. Within $K$ sigma, we have
$$S_t > r t + n - K\sigma\sqrt{t} = r\left(\sqrt{t} - \frac{K\sigma}{2r}\right)^2 + n - \frac{K^2\sigma^2}{4r}$$
When $\displaystyle\;n > \frac{K^2\sigma^2}{4r}\;$, we expect the "return to origin" becomes a $K$ sigma event. This suggests$\color{blue}{^{[1]}}$
$$\lim_{n\to\infty} f_n = 0 \quad\implies\quad A = 0$$
Combine with the constraint $f_0 = 1$, we find $C = 1$ and
$$f_n = \left(\frac{\sqrt{1+4\mu}-1}{2}\right)^n$$
If $\mu > 2$, the $3^{rd}$ root $\displaystyle\;\frac{\sqrt{1+4\mu} - 1}{2} > 1$.
If $C \ne 0$, then $f_n$ will become unbounded for sufficiently large $n$. One again, this contradict with the nature of $f_n$ begin the probability of some event.
This leads to $C = 0, A = 1$ and hence $f_n = 1$.
If $\mu = 2$, the characteristic equation $(*2)$ has a double root at $\lambda = 1$.
The general solution for $(*1)$ now takes the form
$$f_n = A + B \left( -\frac{1+\sqrt{1+4\mu}}{2} \right)^n + C'n$$
One again, it is easy to see $C' = 0, A = 1$ and hence $f_n = 1$.
To summarize:
$$f_n = \begin{cases}
\displaystyle\;\left(\frac{\sqrt{1+4\mu}-1}{2}\right)^n, & \mu < 2\\
1, & \mu \ge 2
\end{cases}$$
Notes
- $\color{blue}{[1]}$ This is a hand waving argument, feel free to justify this rigorously.
Let $L_d = \{1, ..., d\}^2$ be the domain of the random walk. I assume that when OP says
a 2-D random walk that, at any point, has an equal probability of going to any of the adjacent points
they mean that at the boundaries, any "accessible" point is selected with equal probability. Thus, at the point $x = (1, 3)$ and $d = 5$, the points $(1, 2), (1, 4), (2, 3)$ exhaust all the next possible states, each with probability 1/3.
This Markov chain is irreducible and aperiodic on a finite state space, hence ergodic with unique stationary distribution $\pi$. In general, the transition kernel of a finite-state Markov chain can be encoded via a matrix $P$ called the transition matrix. The stationary distribution is then the row vector which satisfies $\pi^T P = \pi^T$ (here the superscript $T$ denotes a transpose). In this case, $P$ would be a $d^2\times d^2$ matrix. You can read more about stationary distributions of Markov chains anywhere you like with Google.
Owing to the asymmetry of the proposal distribution of the Markov chain at the boundary, the stationary distribution will not be uniform. You can see this by running more samples. I ran with $10^6$. Below is a heat map of the probabilities.
Finally, you ask "why" would this be so. If you started from a random location on the grid and ran 1000 steps of the random walk, do you think you could identify with high confidence where you started? No. Each state that's not on the boundary eventually experiences similar "inflow" and "outflow", regardless of where the chain started, so the eventual frequency of visits should be equal for all such states.
Best Answer
Short answer, the average number of steps does not exist.
To show this, suppose that the average (read: expected value) of the number of steps to move 1 step backward exists and is equal to $x$. We can make an equation for $x$ by multiplying the probability of each direction by the expected number of steps based on each direction. If we get a backwards step, then it only takes $1$ step. If we get a forward step, then we have $1$ step, plus the average number of steps to get back to the original position, plus the average number of steps to go backward. This becomes the equation:
$$x=\frac{1}{2}(1)+\frac{1}{2}(1+x+x)$$ $$\Rightarrow x=1+x$$
But this is a contradiction, so there is no finite average number of steps.