How to expand $ {_1}F_1(1;b x;-x)$ in $1/x$? When $x\to\infty$, this fuction reaches the limit value of $\frac{b}{b+1},|b|>1$. But how to get the next term?
I just got an idea.
We can set $ {_1}F_1(1;b;x)=f(b,x)=A(b)+B(b)/x$ and substitute it into the PDE that $f(x)$ satisfies and seek for series solution.
The function $ {_1}F_1(1;b;x)=f(b,x)$ satisfies the follow equation:
$$ x\frac{d}{dx}f(b,x)+f(b,x)=\frac{d}{dx}\left(x\frac{d}{dx}f(b,x)+(b-1)f(b,x)\right)$$
We remark that we can not replace $b$ with $b x$ at this stage. Because the differential operator will work on $b x$ as well if we do so. Only after we did the differentiation can we make the substitution. 
 A: Assuming that the equation is the one in title, change variable $x=\frac 1 y$ to make $$\, _1F_1(1;b x;-x)=\, _1F_1(1;\frac b y;-\frac 1 y)$$ In my opinion, the problem of the next term looks very difficult since $$\, _1F_1(1;\frac b y;-\frac 1 y)=\frac{e^{-1/y} \left(-\frac{1}{y}\right)^{-\frac{b}{y}} (y-b) \left(\Gamma
   \left(\frac{b}{y}-1\right)-\Gamma
   \left(\frac{b}{y}-1,-\frac{1}{y}\right)\right)}{y^2}$$ the development of which as Taylor series around $y=0$ does not look to be trivial.
Looking at the plots of the function for different values of $b$ over the range $0\leq y\leq 1$ effectively reveal almost straight lines. So, since $$\, _1F_1(1; b ;-1)=(-1)^{-b} \left((b-1)\,( !(b-2))-\frac{\Gamma (b)}{e}\right)$$ where appears the subfactorial function and knowing the value $\frac b{b+1}$ when $y=0$ we can make an approximation $$B(b)\approx(-1)^{-b} (b-1)\,( !(b-2))-\frac{(-1)^{-b} \Gamma
   (b)}{e}-\frac{b}{b+1}$$ which is not fantastically good but better than nothing (and, after all, not so bad !).
Instead of using $y=1$, we can use $y=\frac 1e$ and get $$B(b)\approx -\frac{e b}{b+1}-(-1)^{-be } e^{2-e (b+1)} (e b-1) (\Gamma (b e-1)-\Gamma (b
   e-1,-e))$$
We could continue using $y=\frac 1{ke}$ and get $$B(b)\approx e k \left(e^{1-e (b+1) k} (e b k-1) (-k)^{1-e b k} (\Gamma (b e k-1)-\Gamma (b e
   k-1,-e k))+\frac{1}{b+1}-1\right)$$ which really starts to be a nightmare.
For sure $A(b)=\frac{b}{b+1}$.
Have fun !
A: We first prove that 
$$\lim_{x\to\infty}{_1}F_1(1;b x;-x)=\frac{b}{1+b}\tag{1}$$
I am looking for a reference for this proof.
$$
\begin{equation}
\lim\limits_{x\to\infty}{_1}F_1(1;b x;-x)
=\lim\limits_{y\to+\infty}\sum_{k=0}^{\infty}\frac{(1)_k}{k!}\frac{(-x)^k}{(bx)_k}
\end{equation}
$$
$$
\begin{equation}
=\sum_{k=0}^{\infty}\lim\limits_{x\to\infty}\left(\frac{-x}{(bx)}\frac{-x}{(bx+1)}\cdots\frac{-x}{(bx+k-1)}\right)\qquad (\because (1)_k=k!)
\end{equation}
$$
$$
\begin{equation}
=\sum_{k=0}^{\infty}(-1/b)^{k}
=\frac{1}{1+1/b}=\frac{b}{1+b}.\qquad |b|>1\\
\end{equation}
$$
The exchange of the order of summation with limiting is justified because ${_1}F_1(1;b x;-x)$ in (1) is an absolutely convergent series.
Now we let 
$$g(x):=x\left({_1}F_1(1;b x;-x)-\frac{b}{1+b}\right)\tag{2}$$ 
and calculate $\lim_{x\to\infty}g(x)$.
Since
$$
\begin{equation}
\frac{-x}{(bx)}\frac{-x}{(bx+1)}\cdots\frac{-x}{(bx+k-1)}
=(-1/b)^k\left(1-\frac{1}{bx}-\cdots-\frac{k-1}{bx}+O(x^{-2})\right)
\end{equation}
$$
$$
\begin{equation}
=(-1/b)^k\left(1-\frac{k^2-k}{2bx}+O(x^{-2})\right)
\end{equation}
$$
$$
\begin{equation}
\lim_{x\to\infty}g(x)
=\sum_{k=0}^{\infty}\lim\limits_{x\to\infty}x(-1/b)^k\left(-\frac{k^2-k}{2bx}\right)
=\sum_{k=0}^{\infty}(-1/b)^k\left(-\frac{k^2-k}{2b}\right)=-\frac{1}{(1+b)^3}
\end{equation}
$$
Thus
$$
{_1}F_1(1;b x;-x)=\frac{b}{b+1}-\frac{1}{(b+1)^3 x}+O(x^{-2})=\frac{b}{b+1}\left(1-\frac{1}{b(b+1)^2 x}\right)+O(x^{-2})\quad \text{ as } x\to\infty\tag{3}
$$
Numerical results seems to confirm this result.
