Where does this result come from? $$\mathbb{E}\left[\left(\frac{q}{p}\right)^{S_n}\right]\leq \left(\frac{q}{p}\right)^n + \left(\frac{q}{p}\right)^{-n}$$ where $$S_n = \sum_{i=1}^{n}X_i$$ and $0\leq p\leq 1$, $q = 1 - p$?
Edit: The variables $X_i$ take values $1$ with probability $p$ and $-1$ with probability $q=1-p$.
For every $n$, the random variable $(q/p)^{S_n}$ is bounded either by $(q/p)^n$ or by $(p/q)^n$ depending on if $p>q$ or $q<p$. At any case the random variable $(q/p)^{S_n}$ is bounded by the sum of both unless it is infinite, which is the case here. Once again $E[X]\leq \max X$ is used.
Added:
If $q>p$ the expectation is still finite as long as $n$ is finite. Because it is a finite number of summation of finite numbers, which is finite. As $n\to \infty$ right side of the inequality also goes to $\infty$, so there is no problem even asymptotically.
$(p/q)^n$ is calculated for the maximum of $S_n$. For example if $n=3$ then $S_n\leq 3$. If you find the maximum of the left side of your inequality, then it will always be greater than it is for any other case e.g. random walk selects $2$ times $1$ but one time $-1$.