Let $u : \mathbb{R} \rightarrow \mathbb{R}$ be a function such that $\forall x : u'(x) > 0$ and $u'' (x) < 0$.
Is it then true that for any $t > 0$ and $p \in (0, 1)$ the function
$x \mapsto \frac{u^{-1} (p \cdot u(t \cdot x) + (1-p) \cdot u(-x))}{x}$ is decreasing on $(0, \infty)$?
The answer to the question is ``no''. Here is a counterexample.
We set $p=5/6$ and $t=2$. Furthermore, we choose any strictly increasing, strictly concave, and twice continuously differentiable function $u:\mathbb{R}\mapsto\mathbb{R}$ satisfying $$u(-2)=0,\;u(-1)=\frac14,\;u(1)=\frac{107}{144},\;u(2)=\frac56,\;u(4)=1.$$ Because of $$0<\frac14<\frac{107}{144}<\frac56<1$$ and $$\frac{1/4-0}{-1-(-2)}>\frac{107/144-1/4}{1-(-1)}>\frac{5/6-107/144}{2-1}>\frac{1-5/6}{4-2}$$ such a function exists. You want to study the monotonicity of the function $$f(x)=\frac{u^{-1}(p\cdot u(t\cdot x)+(1-p)\cdot u(-x))}x.$$ We have $$f(1)=u^{-1}\big((5/6)\cdot u(2)+(1/6)\cdot u(-1)\big)=u^{-1}(25/36+1/24)=u^{-1}(53/72)$$ and $$f(2)=(1/2)\cdot u^{-1}\big((5/6)\cdot u(4)+(1/6)\cdot u(-2)\big)=(1/2)\cdot u^{-1}(5/6+0)=(1/2)\cdot 2=1.$$ The condition $f(1)<f(2)$ is therefore equivalent to the condition $$\frac{53}{72}<u(1)=\frac{107}{144}.$$ Since this is true, it holds that $f(1)<f(2)$ and the function $f$ is therefore not decreasing.