There is a conceptual error in the OP.
Note that if $w=z(1-z)$, then the condition $\arg(w)\in[0,2\pi)$ restricts the domain of the complex $z$ plane to a half space.
To see this, we write $z=|z|e^{i\arg(z)}$ and $1-z=|1-z|e^{i\arg(1-z)}$ so that $w=|z||1-z|e^{i\left(\arg(z)+\arg(1-z)\right)}$.
Inasmuch as $\arg(z)+\arg(1-z)$ spans a range of $4\pi$ in the complex $z$-plane, then $\arg(w)$ does likewise.
To address the concerns in the OP in detail, we begin with a short primer.
PRIMER:
The complex logarithm $\log(z)$ is defined for $z\ne 0$ as
$$\log(z)=\log(|z|)+i\arg(z) \tag 1$$
It is easy to show that the complex logarithm satisfies
$$\log(z_1z_2)=\log(z_1)+\log(z_2) \tag 2$$
which means that any value of $\log(z_1z_2)$ can be expressed as the sum of some value of $\log(z_1)$ and some value of $\log(z_2)$.
To see that $(2)$ is true, we simply note that $\log(|z_1||z_2|)=\log(|z_1|)+\log(z_2)$ and $\arg(z_1z_2)=\arg(z_1)+\arg(z_2)$.
NOTE: The relationship in $(2)$ is not generally satisfied when the logarithm is restricted on, say, the principal branch for which $\arg(z)=\text{Arg}(z)$, where $-\pi<\text{Arg}\le \pi$.
Using $(2)$, we can write for $z\ne0$, $z\ne 1$
$$\begin{align}
f(z)&=\sqrt{z(1-z)}\\\\
&=e^{\frac12 \log(z(1-z))}\\\\
&=e^{\frac12 \left(\log(z)+\log(1-z)\right)}\\\\
&=e^{\frac12\log(z)}e^{\frac12\log(1-z)}\\\\
&=\sqrt{z}\sqrt{1-z}
\end{align}$$
SELECTING A BRANCH OF $\displaystyle \sqrt{z(1-z)}$
To obtain a specific branch of $\sqrt{z(1-z)}$, we can use a branch of $\sqrt{z}$ and another branch of $\sqrt{1-z}$.
If we select the branches for $\sqrt{z}$ and $\sqrt{1-z}$ to be such that $-\pi<\arg(z)\le \pi$ and $0<\arg(1-z)\le 2\pi$, then the branch of $\sqrt{z(1-z)}$ is such that
$$\sqrt{z(1-z)}=\sqrt{|z||1-z|}e^{i\frac12 (\arg(z)+\arg(1-z))}$$
with $-\pi<\arg(z)+\arg(1-z)\le 3\pi$.
With this choice, it is straightforward to show that $\sqrt{z(1-z)}$ is analytic on $\mathbb{C}\setminus [0,1]$.
EVALUATING THE INTEGRAL
Then, we can write
$$\begin{align}
\oint_C \sqrt{z(1-z)}\,dz&=\int_0^1 \sqrt{x(1-x)}e^{i(0+2\pi)/2}\,dx+\int_1^0\sqrt{x(1-x)}e^{i(0+0)}\,dx\\\\
&=-2\int_0^1 \sqrt{x(1-x)}\,dx \tag 3
\end{align}$$
NOTE: We bypassed consideration of the contributions to the integral from the circular deformations around the branch points since their contributions vansish in the limit as the radii go to zero.
Using Cauchy's Integral Theorem, the value of the integral $\oint_C \sqrt{z(1-z)}\,dz$ is unaltered by deforming $C$ into a circular contour, centered at the origin, of radius, $R>1$. Hence, exploiting the analyticity of $\sqrt{z(1-z)}$ for $R>1$, we have
$$\begin{align}
\oint_C \sqrt{z(1-z)}\,dz&=\oint_{R>1}\sqrt{z(1-z)}\,dz\\\\
&=\int_{-\pi}^{\pi}\sqrt{Re^{i\phi}(1-Re^{i\phi})}\,iRe^{i\phi}\,d\phi\\\\
&=-\int_{-\pi}^{\pi} \left(iRe^{i\phi}\right)^2 \left(1-\frac{1}{Re^{i\phi}}\right)^{1/2}\,d\phi\\\\
&=-\int_{-\pi}^{\pi} \left(iRe^{i\phi}\right)^2 \left(1-\frac{1/2}{Re^{i\phi}}-\frac{1/8}{(Re^{i\phi})^2}-\frac{1/16}{(Re^{i\phi})^3}+O\left(\frac1{(Re^{i\phi})^4}\right)\right)\,d\phi\\\\
&\to -\frac{\pi}{4}\,\,\text{as}\,\,R\to \infty \tag 4
\end{align}$$
Finally, putting together $(3)$ and $(4)$ yields
$$\int_0^1 \sqrt{x(1-x)}\,dx=\frac{\pi}{8}$$
as expected.
NOTE: The expansion leading to $(4)$ is correct given the chosen branches of $\sqrt{z}$ and $\sqrt{1-z}$. Then,
$$\begin{align}
\sqrt{z(1-z)}&=\sqrt{-z^2\left(1-\frac1z\right)}\\\\
&=e^{\frac12\log(-z^2)+\frac12\log\left(1-\frac1z\right)}\\\\
&=iz \sqrt{1-\frac1z}
\end{align}$$
where we used $\log(-1)=i\pi$ and $\log(z^2)=2\log(z)$. Then, upon expanding $\sqrt{1-\frac1z}$ in its Laurent series in the annulus $1<z<\infty$, and setting $z=Re^{i\phi}$, we obtain the expansion used to arrive at $(4)$.
Best Answer
In light of Robert Israel's answer, I would appreciate if someone can point out the flaw in my code.
My simulations suggest the expectation is not zero. With $n=10^5$ samples and $\epsilon=1$, the histogram of $\log \|e_1 + \epsilon (X, Y)\|$ is
The mean gets closer to zero as $\epsilon \to 0$ though, which matches intuition. But the claim does not seem to exactly hold for $\epsilon > 0$.