Is it possible to identify this sequence? Interested by this question, $j$ being a positive integer,
 I tried to work the asymptotics of 

$$S^{(j)}_n=\sum^{n}_{k=0}\frac{\binom{n}{k}}{n^k(k+j)}=\frac{\, _2F_1\left(j,-n;j+1;-\frac{1}{n}\right)}{j}$$ 

I quickly noticed (not a proof) that the asymptotics write

$$S^{(j)}_n=(-1)^j\left(\left(\alpha_0-\beta_0e\right)-\frac{\left(\alpha_1-\beta_1e\right)}{2n}+\frac{\left(\alpha_2-\beta_2e\right)}{24n^2}\right)+O\left(\frac{1}{n^3}\right)$$ 

in which the $\alpha_k$'s and $\beta_k$'s are all positive whole numbers depending on $j$.
What I found is that $$\alpha_0=(j-1)!\qquad \qquad \beta_0=\text{Subfactorial}[j-1]$$ $$\alpha_1=(j+1)!\qquad \qquad \beta_1=\text{Subfactorial}[j+1]$$ $$\alpha_2=(1+3j)(j+2)!$$ but I did not find any formal representation for $\beta_2$.
I give below a list of values ot this last coefficient as a function of $j$.
$$\left(
\begin{array}{cc}
j & \beta_2 \\
 1 & 11 \\
 2 & 60 \\
 3 & 443 \\
 4 & 3442 \\
 5 & 29667 \\
 6 & 281824 \\
 7 & 2936915 \\
 8 & 33374022 \\
 9 & 411167963 \\
 10 & 5462660068 \\
11 & 77886959691 \\
12 & 1186630738810 \\
13 & 19242660629363 \\
14 & 330973762825032
\end{array}
\right)$$
It seems that $\frac{\beta_2}{\text{Subfactorial}[j+2]}$ is close to a straight line but it is not.

Is there any way to identify what is this sequence ?

 A: Notice
$$S_n^{(j)} 
= \int_0^1 \left(1+\frac{x}{n}\right)^n x^{j-1} dx
= \int_0^1 e^{n\log\left(1+\frac{x}{n}\right)} x^{j-1} dx
= \int_0^1 e^{x - \frac{x^2}{2n} + \frac{x^3}{3n^2} + O(n^{-3})} x^{j-1}dx\\
= \int_0^1 e^x \left[1 - \frac{x^2}{2n} + \frac{8x^3+3x^4}{24n^2} + O(n^{-3})\right]x^{j-1} dx
$$
Compare with expansion in question, we get
$$(-1)^j(\alpha_2 - \beta_2 e) = 8A_{j+2}+3A_{j+3}
 $$
where 
$$\begin{align}A_{j} 
&= \int_0^1 e^x x^j dx
= \left(\int_{-\infty}^1 - \int_{-\infty}^0\right) e^x x^j  dx
= \underbrace{\int_0^\infty e^{1-y} (1-y)^j dy}_{x=1-y} - \underbrace{\int_0^\infty e^{-y} (-y)^j dy}_{x = -y}\\
&= (-1)^j\left(e \int_0^\infty e^{-y} (y-1)^j dy - j!\right)
\end{align}
$$
Using the fact 
$$\int_0^\infty e^{-y} (y-1)^n dy = !n \stackrel{def}{=} n! \left(\sum_{s=0}^n \frac{(-1)^s}{s!}\right) = \left\lfloor\frac{n!}{e}+\frac12\right\rfloor$$
is the $n^{th}$ derangement number, we find
$$(-1)^j\left(\alpha_2 - \beta_2 e\right)
= (-1)^j\left\{8[e!(j+2) - (j+2)!] - 3[e!(j+3) - (j+3)!]\right\}\\
\implies
\begin{cases}
\alpha_2 &= 3\times(j+3)! - 8\times(j+2)! = (3j+1)(j+2)!\\
\beta_2  &= 3\,\times\,!(j+3) - 8\,\times\,!(j+2)
\end{cases}
$$
