Let us actually solve and see what exactly happens and what is the reason for this erroneous result you get.
Solving we get $\displaystyle x = \frac{2 \pm \sqrt{4-4(1-\epsilon)}}{2(1-\epsilon)} = \frac{1 \pm \sqrt{\epsilon}}{1-\epsilon} = \frac{1}{1 \mp \sqrt{\epsilon}} = 1 \pm \sqrt{\epsilon} + \epsilon \pm \sqrt{\epsilon^3} + \epsilon^2 \pm \sqrt{\epsilon^5} + \cdots$
When we assume the perturbed solution is $x = \displaystyle \sum_{n\geq 0} c_n \epsilon^n$, we are not accounting for the fractional powers of $\epsilon$.
The problem essentially stems from our inability to represent $\frac{1}{1 \pm \sqrt{\epsilon}}$ as a taylor series about origin.
The error $1=0$ arises when you cancel off $2C_1$ with $-2C_1$. If we try to express $\frac{1}{1 \pm \sqrt{\epsilon}}$ as a taylor series, then you will find that the derivative at the origin $C_1$ is unbounded. Hence, you are canceling of $+\infty$ and $-\infty$ and this is where the error creeps in and hence you get $1=0$.
To avoid this, a better way to deal with the problem would be as follows. If we know that the perturbation is positive, we can replace $\epsilon$ by $\epsilon^2$.
Now the problem becomes $(1-\epsilon^2)x^2 + 2x + 1 = 0$ Solving, we get $x=\frac{1}{1 \mp \epsilon} = 1 \pm \epsilon + \epsilon^2 \pm \epsilon^3 + \epsilon^4 \pm \epsilon^5 + \cdots$