You can continue to play the same "game" that lead to the probabilities for hitting $0$ and $x+y$ and the expectations for $\tau$ and $X_\tau$. Just construct martingales to which you can apply the optional stopping theorem. Note that $X_n$, $X_n^2-n$, $X_n^3-3nX_n$, $X_n^4-6nX_n^2+n(3n+2)$, are all martingales. Applying the optional stopping theorem to the first two yields
$P(X_\tau=0)=\frac{y}{x+y}$, $P(X_\tau=x+y)=\frac{x}{x+y}$, $E(\tau)=xy$, as you already know.
With the next one you can show that
$E(\tau X_\tau)=\frac{1}{3}xy(2x+y)$.
Conditioning on $X_\tau=0$ and $X_\tau=x+y$ gives
$E(\tau X_\tau)=\frac{y}{x+y}E(\tau X_\tau|X_\tau=0) + \frac{x}{x+y}E(\tau X_\tau|X_\tau=x+y)$,
which tells you that
$E(\tau|X_\tau=x+y) =\frac{1}{3}y(2x+y)$, $E(\tau|X_\tau=0)=\frac{1}{3}x(x+2y)$,
i.e. now you know the expectations of the hitting times conditional on player A winning or losing.
Applying the optional stopping theorem to the fourth martingale will give you the second moment of the hitting time...
You can construct martingales that are higher order polynomials in $X_n$ and $n$, or, since after a while this gets a little bit tedious, you might want to put them together and use mgf's, to describe the law of $\tau$ (conditionally on hitting 0 and $x+y$). Just observe that
$E(\exp(\lambda \xi_i))=\cosh(\lambda)$,
so
$M_n(\lambda) = \frac{\exp(\lambda X_n)}{\cosh(\lambda)^n}$
is a martingale.
