This Wikipedia article states that the isomorphism type of a finite simple group is determined by its order, except that:
- $L_4(2)$ and $L_3(4)$ both have order $20160$
- $O_{2n+1}(q)$ and $S_{2n}(q)$ have the same order for $q$ odd, $n > 2$
I think this means that for each integer $g$, there are $0$, $1$ or $2$ simple groups of order $g$.
Do we need the full strength of the Classification of Finite Simple Groups to prove this, or is there a simpler way of proving it?
Although I am not (by any stretch) an expert on finite simple groups, let me flesh out my above comment.
Consider the following QCFSG (i.e., "qualitative" CFSG): with only finitely many exceptions, every finite simple group has prime order, is alternating, or is one of the finitely many known infinite families of Lie type. QCFSG must have been conjectured rather early on, whereas the exact statement of CFSG was much harder to come by, as much of the early work on the classification problem resulted in discovery of new sporadic groups.
I guess that early on someone must have looked at the nonsporadic finite simple groups and noticed that, except for the two exceptions listed above, they have distinct orders. [Assuming this is actually true, that is. I have no reason to doubt it, but I haven't checked it myself.] Once you notice that, if you believe QCFSG, then you certainly think that the order of a simple group determines the group up to finitely many exceptions. It is very hard for me to imagine how you could prove that the number of exceptions is precisely two without knowing the full CFSG.
I cannot resist conveying a story of Jim Milne, whose moral is that you shouldn't feel too bad when you say something absolutely stupid in public: better mathematicians than you or I have said stupider things.
For the source, and for some further fun stories, see
http://www.jmilne.org/math/apocrypha.html