I don't think there is a very conceptual approach giving a meaning to the comparison of conjugacy class sizes, but I think there is a way to make the argument quite short. The first thing to do is to cancel the term $a_1!$ for $i=1$ (for the points remaining fixed) in the denominator against a part of the numerator $n!$, which leaves the numerator equal to $n(n-1)\ldots(a_1+1)$ (the number of factors is the sum of the sizes of the cycles for the conjugacy class, in the group-theoretic sense where cycles cannot have length$~1$).
A first easy simplification rids us the case of mixed cycle structures, classes with is more than one different cycle length${}\geq2$. For such cycle types we can map permutations in the class to another non-trivial cycle type by simply omitting the cycle(s) of the longest length$~l$ from the decomposition. This map commutes with any conjugation by a permutations (which simple relabels elements in the cycles) and is therefore surjective to a single new conjugacy class; moreover it is not injective since there is always more than one way to reconstruct the length$~l$ cycle(s). Therefore such classes are never the smallest non-trivial ones.
The next step is to show that we need only deal with the case of transpositions, possibly flying in formation. More precisely (and with three exceptions that do not occur for $n\geq7$), the class with $m$ disjoint $k$-cycles with $k>2$ is larger than the one with $m$ disjoint $2$-cycles. We have $a_k=m$ and factor our formula as
$$
\frac{n(n-1)\ldots(n-mk+1)}{k^m}\times\frac1{m!},
$$
where the second factor is identical to the one for $k=2$. In the first factor replacing $k$ by $2$ means dropping the last $m(k-2)$ factors from the numerator and dividing the denominator by $(k/2)^m$. Using that a product of $k-2$ distinct positive integers can only be${}\leq k/2$ if $k\leq4$, we see that this decreases the first factor except when $a_1=0$, and $(m,k)\in\{(1,3),(1,4),(2,3)\}$, which happens only for $n=3,4,6$ (it explains that $3$-cycles are least numerous for $n=3$, that for $n=4$ there are as many $4$-cycles as $2$-cycles, namely $6$, and that for $n=6$ there are fewer ($40$) pairs of $3$-cycles then pairs of transpositions $(45)$).
We have reduced to the case of $m$ disjoint transpositions, which we shall compare with isolated transpositions ($m=1$). For fixed $m$, increasing $n$ leaves the denominator unchanged and increases all factors in the numerator by$~1$. This gives more relative increase for $m>1$ than for $m=1$, so if the class with $m$ disjoint transpositions is to be less numerous than that with one transposition, it must already be the case for the minimal possible value of $n$, which is $n=2m$. The formula for the number of products of $m$ disjoint transpositions when $n=2m$ is the product $1\times 3\times\cdots\times(2m-1)$ (which some people, defying the ambiguity, write as $(2m-1)!!$) which is${}\leq\binom{2m}2$ only for $m=2,3$; this explains these cases for $n=4,6$. For $n\geq7$ this does not happen any more.
There are at least two possible proofs, one of them by enumeration and
another one using the exponential formula.
By enumeration, first choose the elements to go on each cycle
(multinomial coefficient):
$$\frac{N!}{\prod_j (j!)^{P_j}}$$
Each of these elements generates $j!/j$ cycles (in placing labels on a
directed cycle all orbits under the action of the cyclic group have
the same size which is $j$):
$$\prod_j \frac{(j!)^{P_j}}{j^{P_j}}$$
Permutations of the size $j$ cycles are not distinguished:
$$\prod_j \frac{1}{P_j!}.$$
This yields the answer
$$\frac{N!}{\prod_j (j!)^{P_j}}
\prod_j \frac{(j!)^{P_j}}{j^{P_j}}
\prod_j \frac{1}{P_j!}
= \frac{N!}{\prod_j j^{P_j} P_j!}.$$
The second proof uses the exponential formula (OGF of the cycle index
of the symmetric group)
$$Z(S_N) = [z^N]
\exp\left(a_1 z + a_2 \frac{z^2}{2} + a_3 \frac{z^3}{3} + \cdots\right).$$
Extracting the coefficient of $[z^N a_1^{P_1} a_2^{P_2} \times\cdots]$
in $N! Z(S_N)$ we get
$$N! [z^N] [\prod_j a_j^{P_j}] \prod_j \exp\left(a_j\frac{z^j}{j}\right)
= N! [z^N] \prod_j \frac{z^{j P_j}}{j^{P_j} P_j!}
\\ = N! [z^N] z^{\sum_j jP_j} \prod_j \frac{1}{j^{P_j} P_j!}
= N! \prod_{j} \frac{1}{j^{P_j} P_j!},$$
the same as in the first version. (Here we take $P_j = 0$ for a cycle
size that is absent.)
Remark. The reason for treating the OGF like an EGF is that we
have the labeled species
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}\textsc{SET}(
\mathcal{A_1}\textsc{CYC}_{=1}(\mathcal{Z})
+ \mathcal{A_2}\textsc{CYC}_{=2}(\mathcal{Z})
+ \mathcal{A_3}\textsc{CYC}_{=3}(\mathcal{Z})
+ \cdots)$$
and hence we are extracting from an EGF.
Best Answer
Here is an attempt at a proof (for all values of $n$), but it is not very elegant I am afraid! Any improvements would be welcome.
Since the number of transpositions is $n(n-1)/2$, we need to trove that the $|C_G(\sigma)|$ divides $2(n-2)!$ for an odd permutation $\sigma \in S_n$.
Odd permutations have an odd number of cycles of even length, so there exists an even $r$ and odd $t$ such that $\sigma$ has exactly $t$ cycles of length $r$. So we can write $\sigma=\sigma_1\sigma_2$, where $\sigma_1$ consists of the cycles of length $r$, and $\sigma_2$ the remaining cycles. Then $C_G(\sigma) = C_{S_{tr}}(\sigma_1) \times C_{S_{n-tr}}(\sigma_2)$, so it is sufficient to prove the result in the case $\sigma = \sigma_1$.
So $n=tr$ and $\sigma$ consists of $t$ cycles of length $r$, and its centralizer $C_G(\sigma)$ is a wreath product $ C_{r} \wr S_t$, which has order $r^tt!$.
Now if $r>4$, then $S_{r-2}$ has a subgroup $H$ of order $r$, and so $S_{t(r-2)}$ has a subgroup $H \wr S_t$ of order $r^tt!$, so this divides $(r-2)t! \le (rt-2)!= (n-2)!$ and we are done.
So we just have to handle the cases $r=2$ and $r=4$.
Write $r^tt! = 2^km$ with $m$ odd. Then $m$ divides $t!$ and $t \le n-2$, so it remains to prove that $2^k$ divides $2(n-2)!$.
Now $2^k$ = $r^t2^j$ where $2^j$ divides $(t-1)!$.
When $r=2$, since $C_r \wr S_{t-1} \le S_{n-2}$, we have $r^{t-1}2^j$ divides $(n-2)!$, so $r^t2^j$ divides $2(n-2)!$ as claimed.
When $r=4$, we have $C_r \wr S_{t-1} \le S_{n-4}$, so $r^{t-1}2^j$ divides $(n-4)!$, and again $r^t2^j$ divides $2(n-2)!$.