Find the equivalent infinitesimal of $1-x_n$

We can show $1- x_n > \frac{\ln n}{2n} \iff x_n < 1-\frac{\ln n}{2n}$. And, this will prove $1-x_n \backsim \frac{\ln n}{n}$ combined with what you've already shown.

Let $f(x) = x^n + x -1$. Then, we want to show $f(1-\frac{\ln n}{2n})>0$. To do so, note that, for any $C>1$, $\ln (1-x) > -Cx$ holds for sufficiently small $x>0$. So, by taking $n$ sufficiently large, we have $\ln (1-\frac{\ln n}{2n})>-C\frac{\ln n}{2n}$.

From this we can obtain a lower bound of $f(1-\frac{\ln n}{2n})$ as follows:

\begin{array} \, f(1-\frac{\ln n}{2n}) & = \exp \{n\ln (1-\frac{\ln n}{2n}) \} - \frac{\ln n}{2n} \\ & > \exp (-nC\frac{\ln n}{2n}) - \frac{\ln n}{2n} \\ & = \frac{1}{n^{C/2}} - \frac{\ln n}{2n} \end{array}

If we choose $1<C<2$ then the last amount is positive for sufficiently large $n$.

Note also that the number '$2$' is nothing special. For any $D>1$ choose $C$ so that $1<C<D$. Then, you can show $f(1-\frac{\ln n}{Dn}) > \frac{1}{n^{C/D}} - \frac{\ln n}{Dn} > 0$ for sufficiently large $n$. This implies there is a $k_n \to 1$ as you desire.