A special case where we can avoid one of the two sums in James Martin's answer is when the "determinant" $p_1p_3-p_2p_4$ of the four probabilities is zero. Then we can treat the random walk as a "sum" of two independent one-dimensional random walks, one taking steps $(+1/2, +1/2)$ or $(-1/2,-1/2)$, the other taking steps $(+1/2,-1/2)$ or $(-1/2,+1/2)$. The original walk is back at the origin precisely when both these 1D-walks are.

If the transition probabilities of the two component walks are $p, (1-p)$ and $q, (1-q)$ respectively, then the expected number of visits to the origin (including the start) is $$\sum_{n=0}^\infty \binom{2n}{n}^2 p^nq^n(1-p)^n(1-q)^n.$$

In terms of the original probabilities, $$pq(1-p)(1-q) = (p_1+p_2)(p_2+p_3)(p_3+p_4)(p_4+p_1).$$

I'm no expert on this type of sum, but according to Maple, $$\sum_{n=0}^\infty \binom{2n}{n}^2 x^n = \frac2{\pi} EllipticK(4\sqrt{x}).$$

The probability of never returning is the reciprocal of this (with $x=pq(1-p)(1-q)$), in other words
$$\frac{\pi}{2EllipticK(4\sqrt{(p_1+p_2)(p_2+p_3)(p_3+p_4)(p_4+p_1)})}.$$

An obvious question is whether this holds also when $p_1p_3 \neq p_2p_4$. Off the top of my head I don't see why it couldn't.