We develop all the necessary theory here without recourse to cycles/orbits/etc.
You are about to enter another dimension. A dimension not only of sight and sound, but of mind. A journey into a wondrous land of imagination.
Next stop, the Transparent-Transposition-Transformation-Tour!
Recall that the symmetric group $S_n$ has $n!$ elements; we assume here that the transformations in $S_n$ operate on the set $\{1,2,\dots,n\}$. We also have
Proposition 1: There is a canonical injection ${\iota_{n+1}^n}: S_n \to S_{n+1}$ sending the bijections in $S_n$ to the bijections in $S_{n+1}$ that keep $n+1$ fixed.
Let ${S_n}^{'} = {\iota_{n+1}^n}(S_n) \subset S_{n+1}$.
Theorem 2: For every bijection $\sigma \in S_{n+1}$ not in ${S_n}^{'}$ there exists a unique $\sigma^{'} \in {S_n}^{'}$ and transposition of the form $(n+1 \, k) \,\text{with}\, 1 \le k \le n$ such that
$\tag 1 \sigma = \sigma^{'} \circ (n+1 \, k)$ Proof
Assume that ${{\sigma}_0}^{'} \circ (n+1 \, k_0) = {{\sigma}_1}^{'} \circ (n+1 \, k_1)$. Now ${{\sigma}_0}^{'} \circ (n+1 \, k_0)$ applied to $k_0$ must be $n + 1$ and ${{\sigma}_1}^{'} \circ (n+1 \, k_1)$ applied to $k_1$ must also be $n + 1$. Since we are dealing with injective mappings, $k_0$ must be equal to $k_1$. But then of course, ${{\sigma}_0}^{'} = {{\sigma}_1}^{'}$.
Since $(n+1)! = (n + 1) n!$, a counting argument shows that these unique representations 'cover' all the permutations in $S_{n+1}$ not in ${S_n}^{'}$. $\quad \blacksquare$
To assist in understanding and increase the 'aesthetics quotient', you can regard the (1) representation as true for any element in $S_{n+1}$, where you allow $k$ to be $0$ and agree that $(n+1 \, 0)$ is the identity transformation.
Corollary 3: Every permutation in $S_{n}$ not equal to the identity transformation $1_{Id}$ can be expressed as the composition of $n-1$ or fewer transpositions.
Proof (Sketch)
Use induction and apply theorem 2. $\quad \blacksquare$
Corollary 4: Let $A: \{2,3,\dots,n\} \to \{1, 2,3,\dots,n\}$ be any function satisfying $A(k) \lt k$. Then the $n-1$ transpositions $(k \, A(k))$ generate the symmetric group $S_n$.
Proof: Exercise.
We are now prepared to answer the OP's question.
Lemma 5: Let $\sigma \in S_{n+1}$ be represented as the composition of $M$ transpositions. Then it can also be expressed as the composition of $M$ transpositions,
$\tag 2 \sigma = \alpha_1 \alpha_2 \cdots \alpha_s \beta_1 \beta_2 \cdots \beta_t \text{ with } M = s + t$ satisfying
$\qquad \text{Each } \alpha_u \in {S_n}^{'}$
And
$\qquad \text{Each transposition } \beta_u \text{ acts on } n + 1$
Proof (Sketch)
Using elementary facts about transpositions (see next section), we can keep 'pushing to the right side' transpositions that act on $n+1$ to new transpositions that still act on $n +1$. $\quad \blacksquare$
The idea now is, see what happens when you try to push $\beta_1$ to the right. The answer is either $M$ remains the same with $s$ increasing by 1 and $t$ decreasing by $1$, or $M$ goes down by $2$ with $t$ decreasing by $2$.
Example 6: Let $n + 1 = 10$ and $\beta_1 = (10 \, 5)$ and $\beta_2 = (10 \, 6)$. Then $\beta_1 \circ \beta_2 = (6 \, 5) \circ (10 \, 5)$.
Example 7: Let $n + 1 = 10$ and $\beta_1 = (10 \, 5)$ and $\beta_2 = (10 \, 5)$. Then $\beta_1 \circ \beta_2$ collapses to the identity transformation.
Lemma 8: Let $\sigma \in S_{n+1}$ be represented as the composition of $M$ transpositions. Then it can also be expressed as the composition of $s + 1$ transpositions,
$\tag 3 \sigma = \alpha_1 \alpha_2 \cdots \alpha_s \beta$ satisfying
$\qquad \text{Each } \alpha_u \in {S_n}^{'}$
And
$\qquad \text{The transposition } \beta \text{ acts on } n + 1$ (or $\beta = \text{Identity}$)
And
$\qquad s + 1 \le M \text{ and } M = s + 1 \text{ mod 2}$ when $\beta \ne \text{Identity}$
And
$\qquad s \le M \text{ and } M = s \text{ mod 2}$ when $\beta = \text{Identity}$
Proof: Exercise.
Theorem 9: Let a permutation $\sigma$ be expressed as the product of $k$ transpositions and also as the product of $j$ transpositions. Then either $k$ and $j$ are both even or they are both odd.
Proof
We shall use induction on the number of elements $\{1,2,\dots,n\}$ that are being permuted.
One can see immediately that the theorem holds for $n = 2$.
Suppose the theorem is true for $n$ and let $\sigma$ be a permutation in $S_{n+1}$ represented as the product of $M$ transpositions. If $\sigma(n+1) = n+1$, then $\sigma \in {S_n}^{'}$ and the argument falls right into a case that can be directly handled by our induction hypothesis. Otherwise, we can apply lemma 8 and put $\sigma$ into the form (3), with $\beta$ not the identity. By our induction hypothesis, the length of the $\alpha\text{-stuff}$ can only be odd or even. Adding $1$ for $\beta$ changes nothing in terms of stating that the parity is well-defined. $\quad \blacksquare$
The OP mentioned these elementary facts about transpositions:
\begin{align}(ab)(ab)&= 1_{Id}\\ (ab)(ac)&=(bc)(ab)\\ (ab)(cd)&=(cd)(ab)\\ (ab)(bc)&=(bc)(ac)\end{align}
With $a = n + 1$, you see how you can 'push to the right' the '$n + 1\text{-transpositions}$', occasionally getting a 'collapse' upon encountering $(ab)(ab)$, since the inverse transformations cancel out.
We prove the following lemma that was analyzed by the OP:
Lemma: If $1_{Id} =\tau_1\tau_2\cdots\tau_M$ where the $\tau$'s are transpositions on $\{1,2,\dots,n+1\}$, then $M$ is even.
Proof
The lemma is true for $n = 0$ and also (if you are skeptical) for $n = 1$.
Assume that the lemma holds for any $n \gt 1$. By lemma 8, for the identity transformation $1_{Id} \in S_{n+1}$ we have a (2) representation,
$\tag 1 1_{Id} = \alpha_1 \alpha_2 \cdots \alpha_s \beta$
respecting the original parity of $M$. Now $\alpha_1 \alpha_2 \cdots \alpha_s$ is in ${S_n}^{'}$. By theorem 2 there is one and only one such representation, so $\beta$ must be the identity (not really there). So we have
$\tag 2 1_{Id} = \alpha_1 \alpha_2 \cdots \alpha_s$
But each $\alpha_u \in {S_n}^{'}$, so by our inductive hypothesis $s$ is even and so $M$ must also be even. $\quad \blacksquare$