Here is a full solution:
This group has only one element of order 1
($a^1 = a$ so if $a^1=1$, then $a=1$).
This group has no elements of order 4 or greater
(4 is outlawed by hypothesis; order 5 or greater implies that $a^1,a^2,a^3,a^4,a^5$ are 5 distinct elements of a 4 element set).
This group has no elements of order 3
(if $a$ has order 3, then $a^2 \neq 1$ and $a^2 \neq a$, so $a^2 =b$ (or $c$, but WLOG we choose $b$). Hence $ab=ba=1$ and $b^2=a$. What about $ca$? $ca=1$ implies $c=b$. $ca=a$ implies $c=1$. $ca=b$ implies $c=a$. $ca=c$ implies $a=1$. Oh no!)
Therefore all elements have order 2, and we finish exactly as you did.
The classification of finite simple groups was one of the great mathematical achievements of the 20th Century. It is also one where a single result on the order of the groups played a key role, namely the Feit–Thompson theorem, or odd order theorem:
Theorem. (Feit-Thompson, 1963) Every group of odd order is soluble*.
The proof is famously long, at 255 pages, and has recently been Coq-verified [1].
The derived subgroup of a soluble group is a proper normal subgroup, and so a soluble group is simple only if it is abelian. Therefore, the Feit-Thompson theorem has the following corollary:
Corollary. Every non-cyclic finite simple group has even order.
There are other results in this vein, with much shorter proofs. For example, Burnside's theorem (Wikipedia contains a proof):
Theorem. (Burnside, 1904) Let $p, q, a, b\in\mathbb{N}$ with $p, q$ primes. Then every group of order $p^aq^b$ is soluble.
Therefore, every non-cyclic finite simple group must have order divisible by three primes. Moreover, at least one of these primes occurs twice in the prime decomposition of the order:
Theorem. (Frobenius, 1893) Groups of square-free order are soluble.
You can find a proof of this theorem on Math.SE here. The answer there links to the article [2], where the theorem is Proposition 17 (page 9). The article also claims that the result is due to Frobenius in [3].
*In American English, solvable.
[1] Gonthier, Georges, et al. "A machine-checked proof of the odd order theorem." International Conference on Interactive Theorem Proving. Springer, Berlin, Heidelberg, 2013.
[2] Ganev, Iordan. "Groups of a Square-Free Order." Rose-Hulman Undergraduate Mathematics Journal 11.1 (2010): 7 (link)
[3] Frobenius, F. G. "Uber auflösbare Gruppen." Sitzungsberichte der Akademie der Wiss. zu Berlin (1893): 337-345.
Best Answer
Here are some major falsehoods implicit in your argument:
(You say "by definition" we get $|ab|={\rm lcm}(m,n)$, but in fact the defition of lcm via universal property is correctly given by $m,n\mid k\Leftrightarrow {\rm lcm}(m,n)\color{Red}\mid k$, not ${\rm lcm}(m,n)=k$.)
Here are some true facts:
As for why it's considered infamously hard (relative to how easy it is to state), I think Herstein's note quoted in Dubuque's answer in the linked question illustrates that well enough.