Finding $F(x)$ from $F(kx),$ where $F(x)$ is the antiderivative of the function $f(x)$. I have that $F(e^{x}x) = e^{x}x^{2} - e^{x}x + e^{x} - 1$, and I would like to find $F(x)$.
Attempt
Since $F(e^{x}x) = e^{x}x^{2} - e^{x}x + e^{x} - 1,$ $F(t) = \alpha_{1}t^{\beta_{1}} + \alpha_{2}t^{\beta_{2}} + \alpha_{3}t^{\beta_{3}} + \alpha_{4}t^{\beta_{4}}.$
Let $t = e^{x}x,$ which means that $F(t) = \alpha_{1}(e^{x}x)^{\beta_{1}} + \alpha_{2}(e^{x}x)^{\beta_{2}} + \alpha_{3}(e^{x}x)^{\beta_{3}} + \alpha_{4}(e^{x}x)^{\beta_{4}} = e^{x}x^{2} - e^{x}x + e^{x} - 1.$
Therefore, $\alpha_{2} = -1, \alpha_{4} = -1, \beta_{2} = 1,$ and $\beta_{4} = 0.$
$F(t) = \alpha_{1}t^{\beta_{1}} - t + \alpha_{3}t^{\beta_{3}} - 1 = e^{x}x^{2} - e^{x}x + e^{x} - 1.$

I am not sure I can understand your question. I hope my answer will help you. We can rephrase your question in the following way:

Let $F$ be a function such that $F(e^{x}x) = e^{x}x^{2} - e^{x}x + e^{x} - 1$ for all $x \in \mathbb{R}$. What is the expression of $F$?

We can start by considering the map $x \mapsto y=x e^x$:

As you can see, your "defining" relation is ambiguous for $x \leq 0$, since as $x \leq 0$, $y$ ranges only from a negative value to $0$, and to each admissible $y$ we associate two distinct values of $x$. The situation is easier if $x>-e^{-1}$, since $x \mapsto x e^x$ is strictly increasing and hence bijective. If $g$ is the inverse of $x \in [-e^{-1},+\infty) \mapsto x e^x$, then 
$$
F(x)=e^{g(x)} g(x)^2 -e^{g(x)}g(x)+e^{g(x)}-1.
$$
Since a closed formula for $g$ is unknown, I doubt this answer will be what you expected. 
To summarize: your identity defines a function $F \colon [-e^{-1},+\infty) \to \mathbb{R}$ in an implicit way. For negative values of $x$ you are in the same situation as this example: given that $G(x^2)= \sin x$, find $G$.
