I am trying to prove that $$ \frac{x^{k}}{{\rm e}^{xk}\int_{x}^{\infty} {\rm e}^{-tk}\,\,t^{k}\,{\rm d}t} = ak + b + o\left(1\right) $$ when $k\to \infty$ for any $x> 1$, or equivalently $$ \int_{0}^{}\left({\rm e}^{-s}\,\, \frac{x + s}{x}\right)^{k}\,{\rm d}s = ck^{-1} + dk^{-2} + o\left(k^{-2}\,\right) $$ where I can derive what should be the correct $a,b,c,d$.
Wolfram seems unable to expand or take limits when functions like $\Gamma(k + 1,kx)$ are involved.
Any idea $?$.
On
We can start by factoring the $x$ out and breaking up the product: $$f_n(x)=\frac{1}{x^n}\int_0^\infty e^{-ns}(x+s)^n\mathrm{d}s$$ Then assuming $n\in\Bbb{N}$ (doesn't matter too much if we're taking $n\to\infty$) we get $$=\frac{1}{x^n}\int_0^\infty e^{-ns} \sum_{k=0}^n \mathrm{C}(n,k)x^{n-k}s^k~\mathrm{d}s=\sum_{k=0}^n \mathrm{C}(n,k)x^{-k}\int_0^\infty s^ke^{-ns}\mathrm{d}s$$
$$=\sum_{k=0}^n \mathrm{C}(n,k)x^{-k} ~\frac{k!}{n^{k+1}}$$ $$=(n-1)!\sum_{k=0}^n \frac{1}{(n-k)!}\cdot\frac{1}{(nx)^k}=(n-1)! \sum_{k=0}^n \frac{1}{k!}\cdot\frac{1}{(nx)^{n-k}}$$
$$=\frac{(n-1)!}{(nx)^{n}} \sum_{k=0}^n \frac{(nx)^k}{k!}\approx \frac{(n-1)!}{(nx)^n}\exp(nx)$$
$$f(x;n)\approx \frac{(n-1)!}{(nx)^n}\exp(nx)\tag{$n\gg 1$}$$ It has occurred to me that we can actually use Stirling's formula $$(n-1)!=\Gamma(n)\approx \sqrt{\frac{2\pi}{n}}\frac{n^n}{e^n}$$ Meaning for large $n$, $$f(x;n)\approx \frac{1}{x^n}\sqrt{\frac{2\pi}{n}}e^{n(x-1)}$$ Maybe this helps?
For $k \in \mathbb{N}$ define $f_k \colon (1,\infty) \to \mathbb{R}$ by $$ f_k (x) = \frac{\mathrm{e}^{k x}}{x^k} \int \limits_x^\infty t^k \mathrm{e}^{-k t} \, \mathrm{d} t \stackrel{t \,= \,x \,+ \,\frac{u}{k}}{=} \frac{1}{k} \int \limits_0^\infty \left(1 + \frac{u}{k x}\right)^k \mathrm{e}^{-u} \, \mathrm{d} u \, .$$ Using the series definition of the exponential function and the monotone convergence theorem we find $$ \lim_{k \to \infty} k f_k(x) = \int \limits_0^\infty \lim_{k \to \infty} \left(1 + \frac{u}{k x}\right)^k \mathrm{e}^{-u} \, \mathrm{d} u = \int \limits_0^\infty \mathrm{e}^{\frac{u}{x}} \mathrm{e}^{- u} \, \mathrm{d} u = \int \limits_0^\infty \mathrm{e}^{- \left(1 - \frac{1}{x}\right)u} \, \mathrm{d} u = \frac{x}{x-1}$$ for $x > 1$ (so $c = \frac{x}{x-1}$ in your notation).
In order to compute the next term(s) of the expansion we write $$ f_k(x) = \frac{1}{k} \int \limits_0^\infty \mathrm{e}^{k \log \left(1 + \frac{u}{k x}\right)} \mathrm{e}^{-u} \, \mathrm{d} u = \frac{1}{k} \int \limits_0^\infty \mathrm{e}^{- \left(1 - \frac{1}{x}\right)u} \exp \left[-\frac{u}{x} \left(1 - \frac{\log\left(1 + \frac{u}{k x}\right)}{\frac{u}{k x}} \right)\right] \, \mathrm{d} u $$ for $k \in \mathbb{N}$ and $x > 1$. This implies $$ k^2 \left[\frac{x}{(x-1) k} - f_k(x)\right] = \int \limits_0^\infty \mathrm{e}^{- \left(1 - \frac{1}{x}\right)u} k \left\{1 - \exp \left[-\frac{u}{x} \left(1 - \frac{\log\left(1 + \frac{u}{k x}\right)}{\frac{u}{k x}} \right)\right]\right\} \, \mathrm{d} u$$ for $k \in \mathbb{N}$ and $x > 1$. Using elementary inequalities for the logarithm and the exponential function we can show that the expression in curly brackets is smaller than $\frac{u^2}{2 k x^2}$, so the dominated convergence theorem allows us to compute \begin{align} \lim_{k \to \infty} k^2 \left[\frac{x}{(x-1) k} - f_k(x)\right] &= \int \limits_0^\infty \mathrm{e}^{- \left(1 - \frac{1}{x}\right)u} \lim_{k \to \infty} k \left\{1 - \exp \left[-\frac{u}{x} \left(1 - \frac{\log\left(1 + \frac{u}{k x}\right)}{\frac{u}{k x}} \right)\right]\right\} \, \mathrm{d} u \\ &= \int \limits_0^\infty \mathrm{e}^{- \left(1 - \frac{1}{x}\right)u} \frac{u^2}{2 x^2} \, \mathrm{d} u = \frac{x}{(x-1)^3} \end{align} for $x > 1$ (so $d = - \frac{x}{(x-1)^3}$ in your notation).
Therefore, for $x > 1$ we have $$f_k (x) \sim \frac{x}{(x-1)k} \left[1 - \frac{1}{(x-1)^2 k} + \mathcal{o} \left(\frac{1}{k}\right)\right]$$ as $k \to \infty$ (the order of the next term is (at least) $\frac{1}{k^3}$, as can be seen by expanding the logarithm and the exponential function further). We see that the condition $x > 1$ is required. In fact, the asymptotic behaviour at $x = 1$ is $$ \lim_{x \to 1^+} f_k (x) = \frac{1}{k} \int \limits_0^\infty \left(1 + \frac{u}{k}\right)^k \mathrm{e}^{-u} \, \mathrm{d} u \stackrel{k \to \infty}{\sim} \sqrt{\frac{\pi}{2 k}} \, , $$ as is shown in this answer to another question.