The problem can be formulated as a special random walk.
Let $X_n$ ($n\ge 1)$ be a set of Bernoulli variables, taking values on $\{1,-1\}$, and let $Y_n=\sum_{i=1}^n X_n$ be our random walk. Making the event $X_n=1$ correspond to incrementing $A_n$, we have:
$$ (A_n-A_0)-(B_n-B_0) = Y_n$$
$$ (A_n-A_0)+(B_n-B_0) = n$$
Hence $A_n = A_0 + (Y_n+n)/2$ and $A_n + B_n = n+A_0+B_0$
Then
$$p_n= P(X_n=1) = \frac{A_{n-1}}{A_{n-1}+B_{n-1}} = \frac{1}{2}\frac{Y_{n-1}+n-1+2A_0}{n-1+A_0+B_0}$$
Letting $\mu_n=E(X_n)$, conditioning on the past, and applying the tower expectation, we get (I spare the details) :
$$ \mu_n= \frac{\alpha +\sum_{i=1}^{n-1}\mu_i}{\beta +n} $$
where $\alpha=A_0-B_0$ and $\beta = A_0+B_0-1$. This has the constant solution: $\mu_n = \frac{\alpha}{\beta+1}=\frac{A_0-B_0}{A_0+B_0}$. Therefore
$$E(Y_n)=n \frac{A_0-B_0}{A_0+B_0}$$
and hence $A_n$ grows (in average) as $A_0 + n (\frac{A_0-B_0}{A_0+B_0}+\frac{1}{2})$
What remains is to compute the variance of $Y_n$, which requires to compute the correlations $E(X_n X_m)$. As a starting point, note that $\sigma_X^2=4 A_0 B_0 /(A_0+B_0)^2$. If $X_n$ were indepdent, $\sigma_Y^2= n \sigma_X^2$ . Because they have positive correlation, one would expect the variance to be larger - but not larger than $n^2 \sigma_X^2$.
Update: A straightforward but tedious computation [*] gives me: $$r=E(X_n X_m) = \frac{s+d^2}{ s(s+1)} \hskip{1cm} n\ne m, \;s=A_0+B_0, \; d=A_0-B_0$$
Then $E(Y_n^2)=n+n(n-1) r$ ; and, because $E(Y_n)=n d/s$ then
$$ {\rm{Var}}(Y_n)=n\frac{s^2-d^2}{s(s+1)}\left(\frac{n}{s} +1\right) \approx n^2 \frac{s^2-d^2}{s^2(s+1)}$$
Going back to $A_n$, we see that its mean grows as
$$\overline{A_n} \approx n \left(\frac{1}{2}+\frac{d}{s}\right)$$
and the variance as
$$ {\rm Var}(A_n) \approx n^2 \frac{s^2-d^2}{4 s^2(s+1)}$$
In particular, $A_0=B_0=1$ ($d=0$, $s=2$) we have
$$\overline{A_n} =1+\frac{n}{2} \hskip{1cm} {\rm Var}(A_n) \approx n^2 \frac{1}{12}$$
If instead $A_0=B_0=100$
$$\overline{A_n} =100+\frac{n}{2} \hskip{1cm} {\rm Var}(A_n) \approx n^2 \frac{1}{804}$$
That is, the behaviour is basically the same in the mean, but the variance is smaller.
[*] I've checked my results against simulations, they seem to agree. BTW, I confess I was surprised to find that both the mean and the correlation of $X_n$ are constants, perhaps I'm missing some basic insight and some simpler solution.