Find stationary distribution for infinite space Markov chain

The condition you give is the correctly identified global balance equation for $\pi, \, P$, however it is much easier to find the closed form of the stationary distribution by proving that it satisfies the detailed balance condition

$$\pi(i) P(i,j) = \pi(j)P(j,i), \qquad \forall \, i,j\, \in S.$$

If we can find such a probability vector $\pi$ which satisfies this condition, then $\pi$ is known to be a stationary distribution for $P$.

In the context of your Markov chain, we know that $P(i,j) = 0$ if $j \neq i \pm 1$ and it suffices to find $\pi$ satisfying $$ \pi(i)P(i,i+1) = \pi(i+1)P(i+1,i)$$ That is $$\pi(i)p = \pi(i+1)(1-p).$$ If we write $\rho = p/(1-p)$ then we have the recurrence relation $$\pi(i+1) = \rho \pi(i) = \cdots = \rho^{i} \pi(1) = \frac1p \, \rho^{i+1} \pi(0),$$ where in the final equality we use that the detailed balance at $i = 0$ is given by $\pi(0) = (1-p) \pi(1)$.

Since the stationary distribution must sum to $1$ (it is a probability distribution) we have

$$1 = \sum_{i=0}^{\infty} \pi(i) = \frac{\pi(0)}p \sum_{i=0}^\infty \rho^i$$ You can now use the geometric series identity to identify when this is well defined, and to find the exact formula for the stationary distribution.

Note: In the above I worked with the method of assuming detailed balance to be true, and then showing that this was in fact the case (for certain $p$, which you will need to check). An alternate approach would have been to first demonstrate that the process is reversible by proving Kolmogorov's Criterion (which shows reversibility, which is equivalent to detailed balance), and then deriving the detailed balance relationship after.

Just write those equations between only two nodes, like you have for $0$. $$1 \cdot \pi_0 = q \cdot \pi_1$$ $$p \cdot \pi_1 = q \cdot \pi_2$$ $$ \ldots $$ $$\pi_{k+1} = \frac{p}{q} \cdot \pi_k = \left(\frac{p}{q}\right)^{k}\cdot \pi_1$$

Then you have $$1 = \sum\limits_{k=0}^{\infty}\pi_k = q \cdot \pi_1 + \pi_1 + \left(\frac{p}{q}\right)^{1}\cdot \pi_1 + \left(\frac{p}{q}\right)^{2}\cdot \pi_1 + \ldots$$

$$1 = \pi_1 \cdot \left( q + 1 + \left(\frac{p}{q}\right)^{1} + \left(\frac{p}{q}\right)^{2} + \ldots\right)$$

$$\pi_1 = \frac{1}{\frac{1}{1 - p/q} + q}$$

you need to be careful, so $p < q$, otherwise no stationary distribution exists.