Let $E_n$ denote the probability of the first return after $n$ steps, and let $P_n$ denote the probability of return after $n$ steps (but not necessarily the first return). Note that $E_n=P_n=0$ if $n$ is odd.

We are interested in $E= E(p)= \sum_{n=1}^{\infty} E_n$. Now note that for $n\ge 1$
$$
P_n = E_n + \sum_{k=1}^{n-1} E_k P_{n-k},
$$
from which it follows that
$$
P = \sum_{n=1}^{\infty} P_n = \sum_{n=1}^{\infty} E_n + \sum_{k=1}^{\infty} E_k \sum_{j=1}^{\infty} P_j,
$$
or in other words $P= E(1+P)$, or
$$
E= \frac{P}{1+P}.
$$
All this is due to Polya, of course.

Now $P_n$ is quite easy to calculate: for odd $n$ it is zero, and for even $n$ it is simply (counting $a$ steps to the right and left, and $b$ steps up and down)
$$
P_n = \sum_{a+b=n/2} \frac{n!}{a!^2 b!^2} \Big(\frac 14\Big)^a \Big(\frac 14\Big)^a \Big(\frac 12p \Big)^b \Big(\frac 12 (1-p)\Big)^b.
$$
We can also express $P_n$ as a double integral:
$$
\frac{1}{(2\pi )^2} \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} \Big( \frac 14 e^{i\alpha} + \frac 14 e^{-i\alpha} + \frac 12p e^{i\beta} + \frac 12 (1-p) e^{-i\beta} \Big)^n d\alpha d\beta.
$$
From this we can see that
$$
1+P= \frac{1}{(2\pi)^2} \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} \frac{2}{2-(\cos \alpha + \cos \beta + i(2p-1)\sin \beta)} d\alpha d\beta.
$$

Note that when $p=\tfrac 12$ this integral diverges, and we recover Polya's theorem that $E=1$. For any given $p<\tfrac 12$ we may clearly calculate the integral. Moreover, it is not hard to get asymptotics from the above, and show that $P$ behaves like $\log (1-2p)^{-1}$ for $p$ near $\tfrac 12$ (forgetting constants here).