The answer is yes: indeed, for any $a>0$ and $b>0$, your equation $(**)$ with any $C_0>0$ determines

a solution $G$ to your ODE $(*)$ such that $G>0$ and $G'<0$ on $(b,\infty)$ -- and even on $(0,\infty)$.

Indeed, rewrite $(**)$ as
\begin{equation}
x=X(G(x)),\quad X(g):=
e^{-2a\left(\text{log}\,g-\frac{1}{g}\right)}\left(C_0+b(2a)^{2a}\text{Gamma}\left(1-2a,\frac{2a}{g}\right)\right).
\end{equation}
Take any $C_0>0$.
Then $X(0+)=\infty$ and $X(\infty-)=0$. Let
\begin{equation}
x_1(g):=X'(g)\,\frac{g^2 e^{2 a \left(\log g-\frac{1}{g}\right)}}{g+1}.
\end{equation}
Then
\begin{equation}
x'_1(g)=-\frac{2 a b e^{-\frac{2 a}{g}} g^{2 a}}{(g+1)^2},
\end{equation}
which is manifestly negative for positive $a,b,g$, so that $X'<0$ on the interval $(0,\infty)$ and

hence the function $X\colon(0,\infty)\to(0,\infty)$ continuously decreases from $\infty$ to $0$ on $(0,\infty)$. So, the inverse function $G:=X^{-1}$ exists and solves the equation $x=X(G(x))$ for $x>0$. Moreover, $G>0$ and $G'=1/(X'\circ G)<0$, as desired.

It remains to verify that this function $G$ is a solution to your ODE $(*)$. To do this, just replace, in $(*)$, $G(x)$ by $g$ and $G'(x)$ by $1/X'(g)$, and then $x$ by $X(g)$, so that $(*)$ is rewritten as $g^2 + 2 a X(g) \frac g{X'(g)} + 2 \frac a{X'(g)}\, (X(g) - b)=0$, which is straightforward to check.

Details of calculations can be seen in the Mathematica notebook
and/or its pdf image.