The relation $a\theta_n^Sb$ iff $(\forall x\in S^n)\, xa=xb$. This is Exercise 1.9.17 of Howie's "Fundamentals of Semigroup Theory".
The Details:
Let $S$ be a commutative semigroup.

Definition 1: Let $\theta_n^S$ for $n\ge 1$ be given by
  $$a\theta_n^Sb \iff (\forall x\in S^n)\, xa=xb.$$
Lemma 1: The relation $\theta_n^S$ is a congruence on $S$ and $$\theta_1^S\subseteq\theta_2^S\subseteq \dots$$

Proof: For all $a\in S$, $xa=xa\forall x\in S^n$. Hence $a\theta_n^Sa$, so $\theta_n^S$ is reflexive.
Suppose $a\theta_n^S b$. Then, for $x\in S^n$, $xa=xb$, i.e., $xb=xa$, so $b\theta_n^Sa$. Hence $\theta_n^S$ is symmetric.
Suppose $a\theta_n^S b$ and $b\theta_n^S c$. Fix $x\in S^n$. Then $xa=xb=xc$, so $a\theta_n^S c$. Hence $\theta_n^S $ is transitive.
Suppose $a\theta_n^S b$ and $c\theta_n^S d$. Then for $x=x_1\dots x_n\in S^n$,
$$\begin{align}
x(ac)&=(xb)c \\
&=(x_1\dots x_{n-1})(x_nb)c \\
&=(x_1\dots x_{n-1})(x_nb)d \\
&=x(bd),
\end{align}$$ so that $ac\theta_n^S bd$. Hence $\theta_n^S $ is compatible.

Let $i\in \mathbb{N}$ and $a\theta_i^S b$. Then for any $x=x_1\dots x_i\in S^i$, $xa=xb$, so for any $c\in S$ we have $(cx_1\dots x_i)a=(cx_1\dots x_i)b$; thus $a\theta_{i+1}^S b$. Hence $\theta_1^S\subseteq\theta_2^S\subseteq \dots$. $\square$

Lemma 2: Let $S$ be a monoid. Then for all $n\in\Bbb N$, $\theta_n^S$ is $1_S$, the identity relation on $S$.

Proof: Let $n\in \Bbb N$. If $a1_Sb$, then by definition $a=b$ and so for all $x$ in $S^n$, $xa=xb$, so that $a\theta_n^S b$ and $1_S\subseteq\theta_S^n$. If $a\theta_n^Sb$, then since $1\in S^n$ we have $a=1a=1b=b$, i.e., $a1_Sb$, so $\theta_n^S\subseteq 1_S$. Hence $\theta_n^S= 1_S$. $\square$

Lemma 3: For $n\in \Bbb N$, denote $S/\theta_n^S$ by $S_n$. Then for $n\ge 2$, $$S_n\simeq S_{n-1}/\theta_1^{S_{n-1}}.$$ 

Proof: User amrsa gives a proof of this (and a proof alone) in an answer below. $\square$
The Question:

Let $S=\langle a\rangle= M(m, r)$ be the monogenic semigroup generated by $a$ with index $m$ and period $r$, where $m>1$. Show that
  $$S/\theta_1^S\simeq M(m-1, r)$$ and deduce that
  $$S/\theta_n^S\simeq M(m-n, r)$$ for $n<m$.
Show also that $S/\theta_n^S$ is isomorphic to the cyclic group of order $r$ for all $n\ge m$.

My Attempt:
(The above proofs are my answers to the subproblems of Exercise 1.9.17, so I'm not approaching you empty-handed.)
I suppose $S/\theta_1^S\simeq M(m-1, r)$ because $\theta_1^S$ simply removes an $a$ from $\langle a\rangle$. I can see it but I can't prove it yet. That $S/\theta_n^S\simeq M(m-n, r)$ comes from Lemma 3 by the same token.
Any help would be appreciated.
 A: This is only about Lemma 3.
By the third isomorphism theorem, 
$$\frac{S}{\theta_n^S} \cong \frac{\frac{S}{\theta_{n-1}^S}}{\frac{\theta_n^S}{\theta_{n-1}^S}},$$
where 
$$\frac{\theta_n^S}{\theta_{n-1}^S} = \left\{ \left\langle \frac{a}{\theta_{n-1}^S}, \frac{b}{\theta_{n-1}^S} \right\rangle : \langle a, b \rangle \in \theta_n^S \right\}.$$
Now,
$$\theta_1^{S_{n-1}} = \theta_1^{S/\theta_{n-1}^S}= \left\{ \left\langle \frac{a}{\theta_{n-1}^S}, \frac{b}{\theta_{n-1}^S} \right\rangle : x \in S/\theta_{n-1}^S \Rightarrow \frac{xa}{\theta_{n-1}^S} = \frac{xb}{\theta_{n-1}^S} \right\},$$
and this can easily be proven to be $\frac{\theta_n^S}{\theta_{n-1}^S}$.

Edit: Proof (?) that it $S \cong M(m,r)$, then $S/\theta_1^S \cong M(m-1,r)$.
I'm using the Wikipedia definition of Monogenic Semigroup.
It seemed to me that $S \cong M(m,r)$ is equivalent to $S = \langle a \rangle$ and $S$ satisfies $$x^{m+r} = x^m,$$
resulting in $S = \{ a, a^2, \ldots, a^m, a^{m+1}, \ldots, a^{m+r-1} \}$.  
Is this sufficient to be $S \cong M(m,r)$?
Let us suppose it is and prove that then, $S/\theta_1^S \cong M(m-1,r)$.
It is enough to prove that
$$\left(\frac{a}{\theta_1^S}\right)^{m-1+r} = \left(\frac{a}{\theta_1^S}\right)^{m-1},$$
which is equivalent to
$$\langle a^{m-1+r}, a^{m-1} \rangle \in \theta_1^S,$$
or yet, equivalent to
$$\forall x \in S ( x \cdot a^{m-1+r} = x \cdot a^{m-1} ).$$
But if $x \in S$ then $x = a^k$, for some $k \geq 1$.
If $k=1$, then this follows from $a \cdot a^{m-1+r} = a^{m+r} = a^m = a \cdot a^{m-1}$;
if $k > 1$, then $x \cdot a^{m-1+r} = a^{k-1} \cdot a^{m+r} = a^{k-1} \cdot a^m = x \cdot a^{m-1}$.
