Probability of reaching a state in asymmetric random walk
Suppose that $S_T &gt; C\ln T$. Let $T_0-1$ be the last time that the random walk reached the origin. Thus if we run an unconstrained biased random walk starting at position $1$ at time $T_0$, we reach beyond $C\ln T$ at time $T$. This can only happen if $T-T_0 \geq C\ln T$.
The probability that an unconstrained biased random walk starting at position $x$ ends up to the right of $x$ after $L$ steps is exponentially small in $L$, by Hoeffding's inequality. In your case, $L \geq C\ln T$, and so this probability is polynomially small in $T$. Taking a union bound over all possible values of $T_0$ (at most $T$ of them), you can choose $C$ so that the error probability is at most $\frac{1}{T^d}$ for $d$ of your choice.
(More formally, choose the $\pm 1$ steps $\xi_1,\ldots\xi_T$ of your walk ahead of time. You start at some fixed $S_0$, which here is implicitly assumed to be $S_0 = 1$, and then $S_{n+1} = |S_n + \xi_{n+1}|$. If $S_T &gt; C\ln T$ then $S_{T_0} + \xi_{T_0+1} + \cdots + \xi_T \geq C\ln T$, by definition of $T_0$. We bound the probability that this happens for fixed $T_0$ and take a union bound.)

As for $\mathbb{E}(S_T)$, since you have a stationary distribution, then your chain is positive recurrent (see Chapter 21 of Markov Chains and Mixing Times). Your chain is not ergodic due to a parity issue, so let's consider the square of your chain (moving two steps at once). The square decomposes into two separate chains. The one of the even integers looks as follows:

At $0$, stay at $0$ with probability $p$, and transition to $2$ with probability $1-p$.
At $2n&gt;0$, transition to $2(n-1)$ with probability $p^2$, stay at $2n$ with probability $2p(1-p)$, and transition to $2(n+1)$ with probability $(1-p)^2$.

The one on the odd integers looks as follows:

At $1$, stay at $1$ with probability $p(2-p)$, and transition to $3$ with probability $(1-p)^2$.
At $2n+1$, transition to $2(n-1)+1$ with probability $p^2$, stay at $2n+1$ with probability $2p(1-p)$, and transition to $2(n+1)+1$ with probability $(1-p)^2$.

Let's focus on the case of even integers. Let $f_{2n} = cn$. Using $p(i,j)$ for the probability to transition from $i$ to $j$,
\begin{align}
&amp;p(0,0) e^{f_0 - f_0} + p(0,2) e^{f_2 - f_0} = p + (1-p) e^{-c}, \\
&amp;p(2n,2n-2) e^{f_{2n-2} - f_{2n}} + p(2n,2n) e^{f_{2n} - f_{2n}} + p(2n,2n+2) e^{f_{2n+2} - f_{2n}} = p^2 e^{-c} + 2p(1-p) + (1-p)^2 e^c.
\end{align}
For any $c &gt; 1$, the first line is strictly less than 1. For $c = 1+\epsilon$, the second line is
$$
p^2 (1 - \epsilon) + 2p(1-p) + (1-p)^2 (1 + \epsilon) + O(\epsilon^2),
$$
which is less than 1 for any small $\epsilon&gt;0$ since $p^2 &gt; (1-p)^2$.
We have thus verified Popov's conditions for geometric ergodicity (see for example Isaacson), and we conclude1 that there exists $\beta &lt; 1$ such that
$$
 \left|\Pr[S_{2T} = 2n \mid S_0 = 0] - \pi^e_{2n}\right| = O(e^{f_0-f_{2n}}\beta^T) = O(\beta^T),
$$
where $\pi^e$ is the stationary distribution of the even component of the squared chain, given by $\pi^e_{2n} = 2\pi_{2n}$. Denoting the corresponding expectation by $E^e = \mathbb{E}(\pi^e)$, we conclude that
$$
 |\mathbb{E}(S_{2T} \mid S_0 = 0) - E^e| = O(\beta^T).
$$
Similarly,
$$
 |\mathbb{E}(S_{2T+1} \mid S_0 = 0) - E^o| = O(\beta^T).
$$
We obtain similar results for any other fixed starting point.
1This equation appears in Popov's paper but not in Isaacson's.