I have found that proof
"Let $\underline{a>1}.$ I assume $a^x$ is continuous, and that the basic exponent law $a^{x+y}=a^xa^y$ holds.
Claim: $a^x$ is convex on $[0,\infty).$ Proof: Because $a^x$ is continuous, it suffices to show $a^x$ is midpoint convex. Suppose $x,y\in [0,\infty).$ Using $(uv)^{1/2} \le (u+v)/2$ for nonnegative $u,v,$ we get $a^{(x+y)/2} = (a^{x} a^{y})^{1/2} \le (a^x+a^y)/2.$
Now if $f$ is convex on $[0,\infty),$ then $(f(x)-f(0))/x$ is an increasing function of $x$ for $x\in(0,\infty).$ This is a simple and easily proved property of convex functions.
Claim: $\lim_{x\to 0^+}(a^x-1)/x$ exists. Proof: All of these quotients are bounded below by $0.$ As $x$ decreases to $0,(a^x-1)/x$ decreases by the above. Because of the lower bound of $0,$ the limit exists.
It follows that $\lim_{x\to 0}(a^x-1)/x$ exists. This follows from the above and the fact that if $x>0,$ then $a^{-x} = 1/a^{x}.$ To handle $0<a<1,$ look at $[(1/a)^x-1]/x$ to see $\lim_{x\to 0}(a^x-1)/x$ exists."
source: zhw, Show $\lim_{h\to 0} \frac{(a^h-1)}{h}$ exists without l'Hôpital or even referencing $e$ or natural log, URL (version: 2017-02-18)
Can someone please explain to me why in the first part a>1 is of relevance.
My second question is how can I use the last sentence to see that the special case delivers everything I want to have?
you need $a >1$ for this part:
For the last sequence:
Note: $$\lim_{h \to 0^-} \frac{a^h-1}{h}=-\lim_{t \to 0^+} \frac{a^{-t}-1}{t}=-\lim_{t \to 0^+} \frac{1-a^t}{t}\frac{1}{a^t}=\lim_{t \to 0^+} \frac{a^t-1}{t}\frac{1}{a^t}$$
Finaly, if $a<1$ write $a=\frac{1}{b}$. Then $b>1$ and $$\lim_{h \to 0} \frac{b^h-1}{h}=-\lim_{h \to 0} \frac{b^t-1}{t}\frac{1}{b^t}$$