Approach 1: to prove $G'(x) < 0$ for all $x > 0$:
Here is a proof using @fedja's idea:
(Actually, it is not difficult without using fedja's idea. But fedja's idea is better.):
We have $G(x) > 0$ for all $x > 0$.
Let
$$f(x) := \ln G(x) = \ln x + \ln \phi(x) - \ln(2\Phi(x) - 1).$$
We have
$$f'(x) = \frac{1}{x} - x - \frac{1}{2\Phi(x) - 1}\cdot 2\phi(x)
= \frac{1 - x^2}{x} - \frac{\mathrm{e}^{-x^2/2}}{\int_0^x \mathrm{e}^{-t^2/2}\,\mathrm{d} t}$$
where we have used
$2\Phi(x) - 1 = 2 \int_0^x \phi(t)\, \mathrm{d} t$.
Using $\mathrm{e}^{-u} \ge 1 - u$ for all $u \ge 0$,
we have $\mathrm{e}^{-x^2/2} \ge 1 - x^2/2 > 1 - x^2$ for all $x > 0$.
Also, we have $\int_0^x \mathrm{e}^{-t^2/2}\,\mathrm{d} t < \int_0^x 1 \,\mathrm{d} t = x$ for all $x > 0$. Thus, we have, for all $x > 0$,
$$f'(x) < \frac{\mathrm{e}^{-x^2/2}}{x} - \frac{\mathrm{e}^{-x^2/2}}{x} = 0.$$
Thus, we have $G'(x) < 0$ for all $x > 0$.
Dealing with $G'(x)$ for the case: $x = 0$:
Now, let us deal with $x = 0$.
Since $\lim_{x\to 0} G(x) = 1/2$, we define $G(0) = 1/2$.
Using L'Hopital rule, we have
$$\lim_{x\to 0} \frac{G(x) - G(0)}{x - 0} = 0$$
that is $G'(0) = 0$ (see Remarks at the end for details).
Approach 2: to prove $G'(x) < 0$ for all $x > 0$:
We have $G(x) > 0$ for all $x > 0$.
Let
$$f(x) := \ln G(x) = \ln x + \ln \phi(x) - \ln(2\Phi(x) - 1).$$
We have
$$f'(x) = \frac{1}{x} - x - \frac{1}{2\Phi(x) - 1}\cdot 2\phi(x).$$
(1) If $x \ge 1$, we have $f'(x) < 1/x - x \le 0$.
(2) If $x\in (0, 1)$, let
$$g(x) := \frac{2\Phi(x) - 1}{1/x - x}f'(x) = 2\Phi(x) - 1 - \frac{x}{1 - x^2}\cdot 2\phi(x).$$
We have
$$g'(x) = 2\phi(x) - \frac{x^2 + 1}{(1 - x^2)^2}\cdot 2\phi(x) - \frac{x}{1 - x^2}\cdot 2\phi(x) \cdot (-x)
= - \frac{2x^2}{(1-x^2)^2}\cdot 2\phi(x) < 0.$$
Also, $\lim_{x\to 0} g(x) = 0$.
Thus, we have $g(x) < 0$ for all $x \in (0, 1)$.
Thus, we have $f'(x) < 0$ for all $x \in (0, 1)$.
Thus, $f'(x) < 0$ for all $x > 0$.
Thus, $G'(x) < 0$ for all $x > 0$.
Remarks: Dealing with $G'(x)$ for the case: $x = 0$::
We have
$$\frac{G(x) - G(0)}{x - 0} =
\frac{\frac{x \phi(x)}{2 \Phi(x) - 1} - \frac12}{x}
= \frac{2x\phi(x) - (2\Phi(x) - 1)}{2x(2\Phi(x) - 1)}.$$
Let $N(x) := 2x\phi(x) - (2\Phi(x) - 1)$
and $D(x) := 2x(2\Phi(x) - 1)$.
We have
$$N'(x) = 2\phi(x) + 2x\phi'(x) - 2\phi(x) = 2x\phi'(x)$$
and
$$D'(x) = 2(2\Phi(x) - 1) + 2x \cdot 2\phi(x).$$
Note that $\lim_{x\to 0} N'(x) = 0$
and $\lim_{x\to 0} D'(x) = 0$.
We have
$$N''(x) = 2\phi'(x) + 2x\phi''(x)$$
and
$$D''(x) = 4\phi(x) + 4\phi(x) + 4x \phi'(x) = 8\phi(x) + 4x\phi'(x).$$
We have $\lim_{x\to 0} N''(x) = 0$
and $\lim_{x\to 0} D''(x) = \frac{8}{\sqrt{2\pi}}$.
We have $\lim_{x\to 0} \frac{N''(x)}{D''(x)} = 0$.
Using L'Hoptial rule, we have $\lim_{x\to 0} \frac{N(x)}{D(x)} = \lim_{x\to 0} \frac{N''(x)}{D''(x)} = 0$.
Best Answer
One remark first: technically $\frac{x\phi(x)}{2\Phi(x)-1}$ is undefined at $x=0$, because plugging in zero leads to an expression of the form $\frac{0}{0}$. To resolve this problem, we note that by l'Hopital's Rule, $$\lim_{x\to 0}\frac{x\phi(x)}{2\Phi(x)-1}=\lim_{x\to 0}\frac{\phi(x)+x\phi'(x)}{2\phi(x)}=\frac{1}{2},$$ (the last equality by direct evaluation) and thus $\frac{x\phi(x)}{2\Phi(x)-1}$ has a continuous extension to all of $\mathbb{R}$. So when $x=0$ we interpret $\frac{x\phi(x)}{2\Phi(x)-1}=\frac{1}{2}$. Okay, with that technicality out of the way, to show that for $x\geq 0$ $$\frac{x\phi(x)}{2\Phi(x)-1}\leq \frac{1}{2}$$ is equivalent to showing that $$x\phi(x)\leq \Phi(x)-\frac{1}{2},$$ and it suffices to show that if $f(x):=x\phi(x)$ and $g(x):=\Phi(x)-\frac{1}{2}$, then $f(0)=g(0)$ and that $f'(x)\leq g'(x)$ for all $x\geq 0$. We have $$f'(x)=\phi(x)+x\phi'(x)\text{ and }g'(x)=\phi(x),$$ so showing $f'(x)\leq g'(x)$ reduces to showing $x\phi'(x)\leq 0$, but this clearly holds because by direct calculation $x\phi'(x)=-x^2\phi(x)\leq 0$. This finishes the proof. Please feel free to comment below if you aren't sure about any of the above; I omitted some details which I thought are easy to verify.