Consider $xe^{x-1}+x-2-\epsilon=0$
Assume the solution can be expanded in terms of $\epsilon$, i.e. $x=a_{0}+a_{1}\epsilon+a_2\epsilon^2+... (1)$
Substitute (1) into the equation, we have
$(a_{0}+a_{1}\epsilon+a_2\epsilon^2+...)e^{(a_{0}+a_{1}\epsilon+a_2\epsilon^2+...)}+(a_{0}+a_{1}\epsilon+a_2\epsilon^2+...)-2-\epsilon=0$
By comparing coefficients, I managed to obtain $a_0=1,a_1=\frac{1}{3},a_2=-\frac{1}{18}$.
My working is quite tedious when I substitute (1) into the Taylor series expansion of $e^x$, especially when collecting the terms involving $\epsilon^2$.
Is there an easier approach to this?
If $\epsilon=0$, the solution would be $x=1$.
So expand as a Taylor series $$xe^{x-1}+x-2=3(x-1)+\sum_{n=2}^\infty \frac {n+1}{n!} (x-1)^n$$ Truncate to some order and use power series reversion.
Truncated to $O\left((x-1)^6\right)$, you will very quicly obtain $$x=1+\frac{\epsilon }{3}-\frac{\epsilon ^2}{18}+\frac{5 \epsilon ^3}{486}-\frac{5 \epsilon ^4}{2916}+\frac{13 \epsilon ^5}{65610}+O\left(\epsilon ^6\right)$$ Use this value and, just by curiosity, expand as a series around $\epsilon=0$ to obtain $$xe^{x-1}+x-2-\epsilon=-\frac{7 \epsilon ^6}{262440}+O(\epsilon ^7)$$