Probability distribution for the position of a biased random walker on the positive integers
For your stable distribution try $$P(X=n) = \left(\frac{p}{q}\right)^n \left(1-\frac{p}{q}\right)$$ which satisfies $P(X=n)= p P(X=n-1) +(1-p-q) P(X=n) + q P(X=n+1)$ and sums to $1$.
If $p=q$ you are in a similar situation to a standard random walk, in that the probability of a return to the origin is $1$, but there is no stable distribution as any initial distribution widens over time.