= Solution
{c}
Let $d=\gcd(a,b)>1$ and put $a'=a/d$, $b'=b/d$. Then $a',b'$ are <coprime integers>, and $p=b'/(a'+b')$. The six requested revisions are as follows.
\b[(b)(i): <communicating classes>.] Every increment is divisible by $d$, so the <residue class> modulo $d$ is preserved. Conversely, if $j-i$ is divisible by $d$, <Bézout's identity> applied to $a',b'$ gives a path just as in (b)(i). Thus the original <Markov chain> is reducible on $\mathbb Z$, with exactly the $d$ closed <communicating classes>
$$
\boxed{r+d\mathbb Z,\qquad r=0,1,\ldots,d-1.}
$$
The specified walk starting at zero stays in $d\mathbb Z$.
\b[(b)(ii): <period of a state in a Markov chain>.] The return equation $au=bv$ reduces to $a'u=b'v$. Hence $u=b'k$, $v=a'k$. For every state, the <period of a state in a Markov chain> is therefore
$$
\boxed{a'+b'=\frac{a+b}{d}.}
$$
This is still at least two, so no state is <aperiodic>.
\b[(b)(iii): <stationary distribution>.] The conclusion is unchanged: there is no <stationary distribution> on $\mathbb Z$. The maximum-mass proof in (b)(iii) applies verbatim, without using <coprimality>. Equivalently, each closed <communicating class>, after translating and dividing its state labels by $d$, is the same centered <random walk> with step sizes $a',-b'$ and has no <stationary distribution>.
\b[(b)(iv): occupation of zero.] On $d\mathbb Z$, the rescaled walk $S_n/d$ has exactly the <transition probabilities> of the coprime walk with parameters $a',b'$. The return-probability argument in (b)(iv) therefore makes this class <null recurrent>. The <renewal process> proof of the occupation limit applies unchanged:
$$
\boxed{\frac1n\sum_{r=0}^{n-1}\mathbf1_{\{S_r=0\}}\longrightarrow0\quad\text{almost surely}.}
$$
Starting in another <residue class> would make every visit to zero impossible, also giving occupation proportion zero.
\b[(c)(i): modified <communicating classes>.] These are unchanged: $C=\{-c-b,\ldots,c+a\}$ is one finite <closed communicating class>, and all other states are transient singleton classes. The original proof used consecutive reset destinations and never used <coprimality>. Resetting to the $b$ consecutive integers $c-b+1,\ldots,c$ connects different <residue classes>, so it would be incorrect to split $C$ according to <residue classes> modulo $d$.
\b[(c)(ii): <limiting distribution>.] The column-sum calculation, the <irreducible Markov chain> property on a finite class, positive self-transition <probabilities>, and geometric entrance argument also required no <coprimality>. Thus for every initial state the same <convergence in distribution> holds:
$$
\boxed{R_n\xrightarrow{\mathrm d}\operatorname{Unif}\{-c-b,\ldots,c+a\}.}
$$
In particular, the denominator $2c+a+b+1$ is unchanged; it is not divided by $d$.
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