The recurrent classes do not affect each other. Once we get into one recurrent class, we never leave, and the structure of the other recurrent class is irrelevant.
Just take an example, say with 4 states and transition probability matrix $(P_{ij})$ given by:
$$ (P_{ij}) = \left[ \begin{array}{cccc}
0 & 1/2 & 1/2 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0
\end{array}
\right] $$
State 1 is transient. State 2 forms an aperiodic recurrent class: If we get to state 2 then we always stay there. States 3 and 4 form a periodic recurrent class: If we get to state 3, we bounce around (periodically) between 3 and 4.
So:
-Given we start in state 2: We have a steady state distribution of $\pi=[0,1,0,0]$ (the limiting probabilities converge to this, and the time average fractions of time also converge to this with probability 1).
-Given we start in state 3: The limiting probabilities do not converge (they oscillate depending on even or odd slots), but the time averages converge to $p = [0,0,1/2,1/2]$ with probability 1.
-Given we start in state 1: Then the time averages will certainly converge, but they will not converge to a constant vector with probability 1. Rather, they will converge to a random vector. What they converge to will depend on the outcome of the first transition. If $p=[p_1,p_2,p_3,p_4]$ are the time averages, then given we start in state 1 we get:
$$ p = \left\{ \begin{array}{ll}
[0, 1, 0, 0] &\mbox{ with prob $1/2$} \\
[0, 0, 1/2, 1/2] & \mbox{ with prob $1/2$}
\end{array}
\right. $$
In general, for a finite state discrete time Markov chain with $K$ recurrent classes, each recurrent class $k \in \{1, \ldots, K\}$ has a unique probability distribution $\pi_k$ that satisfies $\pi_k = \pi_k P$ (where $P$ is the transition probability matrix) and such that $\pi_k$ has support only on the states of recurrence class $k$. If we start at a state in recurrence class $k$, then with probability 1 the time averages converge to $\pi_k$. If we start in a transient state, then the time averages will converge to a random vector $p$. We will eventually end up in one of the recurrent states (being the one we first visit). We can define $\theta_k$ as the probability we first visit recurrence class $k$ (defined for each $k \in \{1, \ldots, K\}$). Then $p$ is a random vector with $p = \pi_k$ with probability $\theta_k$, for $k \in \{1, \ldots, K\}$.
Best Answer
Every state is recurrent because the chain is irreducible and $0$ is recurrent; $0$ is recurrent because no matter where you go from $0$ you will go to some state $i$ after which you will return to $0$ in $i$ additional steps. It follows from irreducibility that the whole chain is also recurrent.
Intuitively, without thinking about abstract Markov chain theory, you can explain the recurrence of the entire chain as follows. Beginning at $i$, you will return to zero, and then you will get a trial with probability $\sum_{j=i}^\infty q_j$ to reach $i$ at some point in the next excursion. If you fail, then you return to zero and get another independent trial with the same success probability. Thus you get infinitely many independent trials with a fixed success probability so there will be a success almost surely.
That being said, this chain can definitely be null recurrent. Indeed the expected return time to $0$ is $\sum_i (i+1) q_i$ (one jump to leave, $i$ to come back), which can easily be infinite for instance if $q_i=\frac{1}{\zeta(\alpha)} \frac{1}{i^\alpha}$ and $1<\alpha \leq 2$.