L’Hospital’s rule applies only to limits of the form $0/0$ and $\infty/\infty$. However, there is a standard trick for converting the indeterminate form $1^\infty$ to one of these forms so that l’Hospital’s rule can be applied, and that’s essentially what you’re doing here. Let me write it up in a way that makes a little clearer exactly what is going on.
Let $L=\displaystyle\lim_{x\to 0}\left(e^x+x\right)^{1/x}$. Then $$\ln L=\ln\lim_{x\to 0}\left(e^x+x\right)^{1/x}=\lim_{x\to 0}\ln\left(e^x+x\right)^{1/x}\;,$$ since $\ln$ is a continuous function. Thus,
$$\ln L=\lim_{x\to 0}\ln\left(e^x+x\right)^{1/x}=\lim_{x\to 0}\frac1x\ln(e^x+x)=\lim_{x\to 0}\frac{\ln(e^x+x)}x\;.$$
This last limit is a $\frac00$ indeterminate form, so l’Hospital’s rule applies:
$$\ln L=\lim_{x\to 0}\frac{\ln(e^x+x)}x=\lim_{x\to 0}\frac{\frac{e^x+1}{e^x+x}}1=\lim_{x\to 0}\frac{e^x+1}{e^x+x}=\frac21=2\;.$$
Finally, then $L=e^{\ln L}=e^2$.
You went astray when you wrote $$\lim_{x \to 0} \frac{\ln\left(e^x+x\right)}{x} = \lim_{x \to 0} \frac{1}{x^2+xe^x}\;:$$ you did not differentiate the numerator correctly, and you did not differentiate the denominator at all.
If $\lim_{x\to a}f(x)=0$ and $\lim_{x\to a}g(x)=1$, then $\lim_{x\to a}\frac{f(x)}{g(x)}=\frac01=0$; there is nothing indeterminate here.
L'Hopital's rule is a local statement: it concerns the behavior of functions near a particular point. The global issues (multivaluedness, branch cuts) are irrelevant. For example, if you consider $\lim_{z\to 0}$, then it's automatic that only small values of $z$ are in play. Saying "take $|z|<1$" is redundant.
Generally, you have a point $a\in\mathbb C$ and some neighborhood of $a$ in which $f,g$ are holomorphic. If $f(a)=g(a)=0$, then
$$\lim_{z\to a}\frac{f(z)}{z-a}=f'(a),\qquad \lim_{z\to a}\frac{g(z)}{z-a}=g'(a) \tag{1}$$
hence
$$\lim_{z\to a}\frac{f(z)}{g(z)}= \lim_{z\to a}\frac{f(z)/(z-a)}{g(z)/(z-a)} =\frac{f'(a)}{ g'(a)}$$
Note that the above is a simple special case of the L'Hopital's rule, because we have (1). It's basically just the definition of derivative.
Best Answer
Hint: use that $$f(x)\to 0\implies\lim\frac{e^{f(x)}-1}{f(x)}=1.$$