This is the case of incomplete gamma function
$\gamma(s,x)$ but you can apply the case to more general function $f(s,x)$. Is possible to have the case result integral function only dependent only from s, deleting x?
Note the following formula has been changed by suggestion of the first answer:
$$
\int_{0}^{1} e^{-xt} t^{s-1} dt = x^{-s} \gamma(s,x)= x^{-s} P(s)
$$
then $ \gamma(s,x)= \frac{g(x)}{g(x)} P(s)$.
And if not correct can it be any case $\gamma(s)$
I am having a hard time understanding exactly what you are asking but believe this will clear it up. By definition $$ \gamma(s,x):=\int_0^x t^{s-1}e^{-t}\,\mathrm dt. $$ Now consider the integral you presented: $$ I(s,x)=\int_0^xt^{s-1}e^{-xt}\,\mathrm dt. $$ Substituting $u=xt$ we are able to write $$ I(s,x)=\int_0^1(u/x)^{s-1}e^{-u}\,\frac{\mathrm du}{x}=x^{-s}\int_0^xu^{s-1}e^{-u}\,\mathrm du=x^{-s}\gamma(s,1). $$ Hence, $$ \int_0^xt^{s-1}e^{-xt}\,\mathrm dt=x^{-s}P(s), $$ with $P(s)=\gamma(s,1)$. This shows you had an error in the calculations provided in your post.