By using some more powerful results, it is possible to do this a lot easier.
The two main ingredients for this will be the following:
Burnside's $pq$-Theorem: If only two distinct primes divide the order of $G$, then $G$ is solvable.
Burnside's Transfer Theorem: If $P$ is a $p$-Sylow subgroup of $G$ and $P\leq Z(N_G(P))$ then $G$ has a normal $p$-complement.
So when looking for a non-abelian simple group, the first result immediately tells us that we can assume at least $3$ distinct primes divide the order of $G$.
The way we will use the second result is the following:
Let $P$ be a $p$-Sylow subgroup where $p$ is the smallest prime divisor of $|G|$. We will show that if $|P| = p$ then $P\leq Z(N_G(P))$ (and then the above result says that $G$ is not simple).
To see this, we note that $N_G(P)/C_G(P)$ is isomorphic to a subgroup of $\rm{Aut}(P)$ (this is known as the N/C-Theorem and is a nice exercise), which has order $p-1$, and since the order of $N_G(P)/C_G(P)$ divides $|G|$, this means that $N_G(P) = C_G(P)$ and hence the claim (since we had picked $p$ to be the smallest prime divisor).
The same argument actually shows that if $p$ is the smallest prime divisor of $|G|$, then either $p^3$ divides $|G|$ or $p = 2$ and $12$ divides $|G|$. The reason for this is that if the $p$-Sylow is cyclic of order $p^2$ the precise same argument carries through, since in that case all prime divisors of $|\rm{Aut}(P)|$ are $p$ or smaller.
If $P$ is not cyclic then the order of $\rm{Aut}(P)$ will be $(p^2-1)(p^2 - p) = (p+1)p(p-1)^2$ and the only case where this can have a prime divisor greater than $p$ is when $p = 2$ in which case that prime divisor is $3$, and we get the claim.
In summary we get that the order of $G$ must be divisible by at least $3$ distinct primes, and either the smallest divides the order $3$ times, or the smallest prime divisor is $2$ (actually, by Feit-Thompson, we know this must be the case, but I preferred not to also invoke that), and $3$ must also divide the order.
This immediately gives us $60$ as a lower bound on the order of $G$, and the next possible order would be $2^2\cdot 3\cdot 7 = 84$ and all further orders are greater than $100$. So we are left with ruling out the order $84$ (which you have done), and showing that the only one of order $60$ is $A_5$.
You are wrong, $\mathbb Z_{100}$ is an option. But it appears twice in your list: Note that $4$ and $25$ do not share any prime factors, so by the Chinese Remainder Theorem,
$$\mathbb Z_4 \times \mathbb Z_{25} \cong \mathbb Z_{100}.$$
As pointed out in the comments, an elementary way to see this is as follows. The element $(1, 1) \in \mathbb Z_4 \times \mathbb Z_{25}$ has order $100$, because if $n$ is a multiple of $4$ and a multiple of $25$, then it is a multiple of $\text{lcm}(4, 25) = 100$. So $(n, n)$ is zero in $\mathbb Z_4 \times \mathbb Z_{25}$ for no $n$ below $100$. Thus, there are at least $100$ elements in this group.
Best Answer
If $n$ is the order of a nonabelian finite simple group then $n$ has at least 3 prime factors (Burnside's $p^a q^b$ theorem) and Sylow's theorem implies that the number $n_p$ of Sylow $p$-subgroups is a nontrivial divisor of $n / p^{v_p(n)}$ congruent to $1$ mod $p$.
Moreover, Burnside's transfer theorem implies
Proof: Let $P \leq G$ be a Sylow $p$-subgroup. Assuming $v_p(n) \leq 2$, $P$ is abelian. By Burnside's transfer theorem (see the answer at https://math.stackexchange.com/a/484777/23805 for discussion), $N_G(P)$ acts nontrivially on $P$. Since $P$ is abelian, $P \leq C_G(P)$, so $N_G(P)/C_G(P)$ is isomorphic to a nontrivial subgroup of $\operatorname{Aut}(P)$ of $p'$-order. Hence $|N_G(P)| = n / n_p$ and $|\operatorname{Aut}(P)|$ share a prime factor other than $p$. Now consider the possibilities for $P$ and $\operatorname{Aut}(P)$. $\square$
Trawling through all odd $n \leq 10000$ (I recommend using technology) shows that we have reduced to three possibilities:
Probably @DerekHolt can rule these out using Sylow's theorem in clever ways.