I see more in your proof that is wrong then is right. For example equation (1) is actually equal to 2. Equation (2) you are claiming that 2 < 1.something, so is also wrong. In equation 3 you lost a 9 somewhere.

Your proof is so riddled with notation and math errors that it is pretty much meaningless.

There's no need for voodoo. We just take the n limit first to get a function of m and then take the m limit.
[itex]L\{f(x)\}= \lim_{m->oo} \lim_{n->oo} \frac{1- \left( exp(-(s+ \alpha )m/n) \right)^n}{1-exp(-(s+ \alpha )m/n)} \frac{m}{n}
= \lim_{m->oo} \lim_{n->oo} \frac{1- \left( exp(-(s+ \alpha )m) \right)}{1-1+(s+ \alpha )m/n+O(n^{-2})} \frac{m}{n}[/itex]

[itex]

= \lim_{m->oo} \lim_{n->oo}m \frac{1- \left( exp(-(s+ \alpha )m) \right)}{(s+ \alpha )m+O(n^{-1})}

= \lim_{m->oo} \lim_{n->oo}\frac{1- \left( exp(-(s+ \alpha )m) \right)}{(s+ \alpha )+O(n^{-1})}

= \lim_{m->oo} \frac{1- \left( exp(-(s+ \alpha )m) \right)}{(s+ \alpha )}

= \frac{1}{(s+ \alpha )}
[/itex]
Provided that [itex](s+ \alpha )>[/itex], which is exactly the same condition as is needed for the proper integral to have a finite value.