This conjecture is true.

Indeed, letting $r:=\rho$, $a:=\gamma$, $t:=rx$, and $u:=e^r-1$, we see

\begin{equation*}
f(x)=R(t):=\frac{e^t-1}{(1+u)e^{-a t}-e^{t-a t}+e^t-1}. \tag{10}\label{10}
\end{equation*}
So, the problem can be restated as follows:

Show that for each real $u>0$ there is some real $a_u>0$ such that for each $a\ge a_u$ there is some $t_{u,a}\in(0,\ln(1+u))$ such that $R''\ge0$ on $[0,t_{u,a}]$ and $R''\le0$ on $[t_{u,a},\ln(1+u)]$.

In what follows, it is assumed that $u>0$, $a$ is a large enough (depending on $u$) positive real number, and $t\in(0,\ln(1+u)]$ -- unless otherwise specified.
Note that $(1+u)e^{-a t}-e^{t-a t}+e^t-1>e^{-a t}-e^{t-a t}+e^t-1=(e^t-1)(1-e^{-a t})>0$, so that \eqref{10} makes sense.

We have
\begin{equation*}
\begin{aligned}
g(t)&:=R''(t)e^{-a t} \left(-e^{a t}+e^{a t+t}-e^t+u+1\right)^3 \\
&=-a^2 (u+1) e^{a t}+e^{a t+2 t} \left(-\left(a^2 (u+3)\right)-2 a
u-u\right) \\
&+e^t (u+1) \left(a^2 (u+3)+2 a u+u\right)+e^{2 t}
\left(-\left(a^2 (2 u+3)\right)-2 a u+u\right) \\
&+e^{a t+t}
\left(a^2 (2 u+3)+2 a u-u\right)+a^2 e^{3 t}+a^2 e^{a t+3
t}-a^2 (u+1)^2.
\end{aligned}
\end{equation*}
Note that $-e^{a t}+e^{a t+t}-e^t+u+1>-e^{a t}+e^{a t+t}-e^t+1=(e^t-1)(e^{a t}-1)>0$, so that $g(t)$ equals $R''(t)$ in sign. So, the problem can be further restated as follows:

Show that for each real $u>0$ there is some real $a_u>0$ such that for each $a\ge a_u$ there is some $t_{u,a}\in(0,\ln(1+u))$ such that $g\ge0$ on $[0,t_{u,a}]$ and $g\le0$ on $[t_{u,a},\ln(1+u)]$.

To prove this, kill the exponentials of the form $e^{ct}$ for constant $c$'s one by one, by letting

\begin{equation*}
D(t):=\frac{g'(t)}{e^t},\quad D_2(t):=\frac{D'(t)}{e^t},\quad D_3(t):=\frac{D_2'(t)}{e^t},
\end{equation*}
\begin{equation*}
D_4(t):=\frac{D_3'(t)}{e^{(a-3)t}},\quad D_5(t):=\frac{D_4'(t)}{e^t},\quad D_6(t):=\frac{D_5'(t)}{e^t}.
\end{equation*}
Then
\begin{equation*}
D_6(t)=6 a^3 (a+1) (a+2) (a+3) e^t \\
-2 a (a+1) \left(a^2+a-2\right)
\left(a^2 (u+3)+2 a u+u\right),
\end{equation*}
which is clearly increasing in $t$, from $D_6(0)=-2 a (a+1) (a+2) \left(a^3 u+a^2 (u-12)-a u-u\right)<0$ eventually (that is, for all large enough $a$, depending on $u>0$) and $D_6(\ln(1+u))=2 a (a+1) (a+2) \left(2 a^3 u+4 a^2 (2 u+3)+a u+u\right)>0$ eventually (and, in fact, for all real $a>0$ and $u>0$).

So, eventually $D_6$ is $-+\,$: that is, for some $t_{6;a,u}\in(0,\ln(1+u))$ we have $D_6<0$ on $[0,t_{6;u,a})$ and $D_6>0$ on $(t_{6;u,a},\ln(1+u)]$.

So, eventually $D_5$ is down-up -- that is, for some $t_{6;a,u}\in(0,\ln(1+u))$ we have $D_5$ is decreasing
on $[0,t_{6;u,a}]$ and increasing on $[t_{6;u,a},\ln(1+u)]$. Also, eventually
$D_5(0)=-a (a+1) \left(10 a^3 u+a^2 (5 u-36)-13 a u-2 u\right)<0$ and $D_5(\ln(1+u))=a (a+1) \left(a^4 u^2+a^3 u (9 u+14)+a^2 \left(16 u^2+43
u+36\right)+a u (6 u+13)+2 u (2 u+1)\right)>0$. So, eventually $D_5$ is $-+$.

So, eventually $D_4$ is down-up. Also, eventually
$D_4(0)=2 a^2 \left(-11 a^2 u+6 a (u+2)+5 u\right)<0$ and $D_4(\ln(1+u))=a (u+1) \left(2 a^4 u^2+7 a^3 u (u+2)+8 a^2 \left(u^2+3
u+3\right)+5 a u (u+2)+2 u^2\right)>0$. So, eventually $D_4$ is $-+$.

So, eventually $D_3$ is down-up. Also, eventually
$D_3(0)=-a \left(14 a^2 u+3 a (3 u-4)+u\right)<0$ and $D_3(\ln(1+u))\frac{(1+u)^2}{a}=(u+1)^a\left(a^3 u^2+2 a^2 u (u+2)+a \left(-u^2+3 u+6\right)-u (2
u+1)\right) +6 a (u+1)^2>0$. So, eventually $D_3$ is $-+$.

So, eventually $D_2$ is down-up. Also, eventually
$D_2(0)=-12 a (a+1) u<0$ and $D_2(\ln(1+u))(1+u)^2
=(u+1)^a\left(-a^2 u (u+1) (3 u+2)-a u (u+1) (7 u+8)-2 u
(u+1)^2\right) +2 a^2 u (u+1)^2-4 a u (u+1)^2+2 u
(u+1)^2<0$. So, eventually $D_2<0$.

So, eventually $D$ is decreasing. Also, eventually
$D(0)=u \left(a^2 u+2 a (u-3)+u\right)>0$ and $D(\ln(1+u))(1+u)^2
=(u+1)^a\left(-a^2 u^2 (u+1)^2-a u (5 u+4) (u+1)^2-u (2 u+3)
(u+1)^2\right) -2 a u (u+1)^3+3 u (u+1)^3<0$. So, eventually $D$ is $+-\,$: that is, for some $t_{1;a,u}\in(0,\ln(1+u))$ we have $D>0$ on $[0,t_{1;u,a})$ and $D<0$ on $(t_{1;u,a},\ln(1+u)]$.

So, eventually $g$ is up-down -- that is, for some $t_{1;a,u}\in(0,\ln(1+u))$ the function $g$ is increasing on $[0,t_{1;u,a})$ and decreasing on $(t_{1;u,a},\ln(1+u)]$. Also, eventually
$g(0)=(2 a+1) u^2>0$ and $g(\ln(1+u))(1+u)^2
=(u+1)^a\left(-2 a u^2 (u+1)^3-u (u+2) (u+1)^3\right) +2 u
(u+1)^4<0$.

So, eventually $g$ is $+-$. $\quad\Box$