Generating function of the Confluent Hypergeometric Function of the First Kind Let $x>0$ and $t\in(-1,1)$. Consider
$$\sum_{m=1}^\infty t^m \sum_{l=1}^m\binom{m-1}{l-1}\frac{(-x)^l}{l!}\,.$$
Can you find a closed expression for this series? Thank you for your time!
It reminds me the generating function associated with the generalized Laguerre polynomials $L_{m}^{(\alpha)}(x)$ with $\alpha=-1$. In addition, maybe it can be useful to note that the series can be expressed as
$$-x \sum_{m=1}^\infty t^m\, {}_1\!F(1-m;2;x)\,,$$
where I have introduced the "Confluent Hypergeometric Function of the First Kind" ${}_1\!F(a;b;c)$.
 A: For $0 < t < 1/2$ it seems to be equal to
$$
\boxed{e^{xt/(t-1)} - 1}.
$$
The proof is a bit tedious, maybe someone will propose something better later.
Consider
$$
S = \sum_{m=1}^{\infty} t^m \sum_{l=1}^m \binom{m-1}{l-1} \frac{(-x)^l}{l!}.
\tag{1}
$$
I will use integral representation of binomial coefficient, i.e.
$$
{n\choose k} = 
\frac{1}{2\pi i}
\oint_{|z|=\rho} \frac{(1+z)^n}{z^{k+1}} \; dz.
\tag{IR}
$$
Taking (IR) into account (1) transforms into
$$
S = \sum_{m=1}^{\infty} t^m \sum_{l=1}^m \frac{1}{2\pi i} \oint_{|z|=1} \frac{(1+z)^{m-1}}{z^l} \; dz \frac{(-x)^l}{l!},
\tag{2}
$$
as an integration contour I choose a circle radius $1$, counterclockwise.
Now we may note the following "overflow equality"
$$
l > m \rightarrow \oint_{|z|=1} \frac{(1+z)^{m-1}}{z^l} \; dz = 0.
\tag{OE}
$$
With (OE) we can extend the sum in (2) to infinity $\sum_{l=1}^m \to \sum_{l=1}^\infty$ and move everything into the integral
$$
S = \frac{1}{2\pi i} \oint_{|z|=1} 
\underbrace{\sum_{m=1}^{\infty} t^m (1+z)^{m-1}}_{t/(1-t-tz)}
\underbrace{\sum_{l=1}^\infty  \frac{(-x)^l}{z^l l!}}_{e^{-x/z} - 1}\; dz,
\tag{3}
$$
which yields after simplification
$$
S = \frac{1}{2\pi i}\oint_{|z|=1} \frac{e^{-x/z} - 1}{(1-t)/t-z} \; dz.
\tag{4}
$$
For convenience I perform variables change in (4) $z \to 1/\xi$, one may note that integration contour still remains unit circle,but the direction is clockwise now
$$
S = \frac{t}{2\pi i (1-t)}\oint_{|\xi|=1} \frac{e^{-x\xi} - 1}{\xi(\xi- t/(1-t))}\; d\xi.
\tag{5}
$$
For $0 < t < 1/2$ this function has $2$ poles inside the integration contour ($\xi = 0$, $\xi = t/(1-t)$), using residues we get the result
$$
S = e^{xt/(t-1)} - 1.
\tag{R}
$$
As a simple test one can use Wolfram Mathematica
fnum[t_, x_] := 
 NSum[t^m*Sum[Binomial[m - 1, l - 1]*(-x)^l/l!, {l, 1, m}], {m, 1, 
   1000}]
fan[t_, x_] = Exp[t*x/(t - 1)] - 1

Comparison for $t= 0.1,0.2,0.3,0.4$ follows

A: In addition to the excellent answer by @guest, it is also possible to use the generating function for the generalized Laguerre polynomials
\begin{equation}
 (1-t)^{-\alpha-1}\exp\left(\frac{xt}{t-1}\right)=\sum_{n=0}^{\infty}L^{(\alpha%
)}_{n}\left(x\right)t^{n} \text{ when } |t|<1
\end{equation}
as guessed in the OP. From the representation of the series in terms of confluent hypergeometric function
\begin{equation}
 S(t,x)=-x \sum_{m=1}^\infty t^m\, {}_1F_1(1-m;2;x)
\end{equation}
and using the generalized Laguerre polynomials representation
\begin{equation}
 L^{(\alpha)}_{n}\left(x\right)=\frac{{\left(\alpha+1\right)_{
n}}}{n!}{{}_{1}F_{1}}\left({-n\atop\alpha+1};x\right)
\end{equation}
with $\alpha=1,n=m-1$, it comes
\begin{align}
 S(t,x)&=-x \sum_{m=1}^\infty t^m\frac{(m-1)!}{{\left(2\right)_{
m-1}}}L_{m-1}^{(1)}(x)\\
&=-x \sum_{m=1}^\infty \frac{t^m}{m}L_{m-1}^{(1)}(x)\\
\end{align}
Then, using the generating function above
\begin{align}
 \frac{\partial S(t,x)}{\partial t}&=-x\sum_{m=1}^\infty t^{m-1}L_{m-1}^{(1)}(x)\\
 &=-x\frac{\exp\left(\frac{xt}{t-1}\right)}{(1-t)^2}\\
 &=\frac{\partial}{\partial t}\left[ \exp\left(\frac{xt}{t-1}\right)\right]
\end{align}
With $S(0,x)=0$ we have
\begin{align}
 S&=\int_0^t \frac{\partial S(\tau,x)}{\partial \tau}\,d\tau\\
 &= \exp\left(\frac{xt}{t-1}\right)-1
\end{align}
