I asked this question a while ago, but got no answer. Eventually I answered it myself, but I'm not completely sure whether my answer is correct or not. I don't know if it is correct to post another question about it, but I'd like my proof to be verified. So here it is:
We want to show that $\forall \epsilon \gt 0 : \exists \delta \gt 0 : ||(x,y)||< \delta \implies |\frac{\ln(1-x-y)}{x+y}+1| \lt \epsilon$.
Fix $\epsilon \gt 0$. We know that $\lim_{\phi \to 0} \frac{ln(1+\phi)}{-\phi} = -1$. So there's $\delta_1 \gt 0$ such that $|\phi| \lt\delta_1 \implies |\frac{ln(1+\phi)}{-\phi}+1| \lt \epsilon$. Let $\delta = \delta_1$ and $\xi = -x-y$. Suppose $||(x,y)|| \lt \delta $. Since all norms in $R^n$ are equivalent we can use the norm of the sum. Then:
$||(x,y)|| = |x|+|y| \geq |x+y| = |\xi|$. Since $|\xi| < \delta_1 $, then $|\frac{ln(1+\xi)}{-\xi}+1| < \epsilon$ therefore $|\frac{ln(1-x-y)}{x+y}+1| < \epsilon $. So $\lim_{(x,y) \to (0,0), x+y \neq 0}{\frac{\ln(1-x-y)}{x+y} } = -1$ Q.E.D.
Your proof is fine.
If you look in more detail, then essentially you only used the fact that $\frac{\ln(1-x-y)}{x+y}$ is a function of just $x+y$, and then used this as a new variable along with the one-variable limit.
The proof you describe would then also be the "ideal" way of copying the $\epsilon-\delta$ from the one to the two dimension case i.e. via change of variable.