Is there a real sequence $0 \leq a_n $ such that for $Re(z) > 1 $ we have :
$$\ln(z) = a_0 + \sum_{n=1}^{\infty} (-1)^n \cdot a_n \cdot z^{1/n}$$
Or
$$\ln(z) = a_0 + \sum_{n=1}^{\infty} (-1)^{1+n} \cdot a_n \cdot z^{1/n}$$
where we take the principle logarithm for $\ln(z)$
Is that sequence $a_n$ unique ? Does that sequence $a_n$ have any closed forms ?
Let $z=re^{i\theta}$.
On left hand side, we know: $$\ln(z)=\ln(re^{i\theta})=\ln(r)+i\theta$$
On right hand side, as you requested(I changed it a bit for simplicity): $$a_0 + \sum_{n=1}^{\infty} a_n \cdot z^{1/n}$$ $$=a_0 + \sum_{n=1}^{\infty} a_n \cdot r^{1/n}e^{i\theta/n}$$ $$=\sum_{n=1}^{\infty} a_n \cdot r^{1/n}\cos\left(\frac{\theta}{n}\right)+i\sum_{n=1}^{\infty} a_n \cdot r^{1/n}\sin\left(\frac{\theta}{n}\right)$$
By comparing real and imaginary parts respectively, we obtain:
$$\ln(r)=a_0 + \sum_{n=1}^{\infty} a_n \cdot r^{1/n}\cos\left(\frac{\theta}{n}\right)$$ $$\theta = \sum_{n=1}^{\infty} a_n \cdot r^{1/n}\sin\left(\frac{\theta}{n}\right)$$
However, fundamentally, we know that $r$ and $\theta$ in general is independent of each other. Therefore, $a_n=0$ for $n \ge 1$ to remove dependence. Then we have $$\ln(r)=a_0 \rightarrow r=e^{a_0}$$ and $$\theta=0$$. But $r$ and $\theta$ are variables, not constants!
Thus, we concluded your request is impossible.