Generalized Laplace distribution? definition:
Laplace distribution $Lap(\mu, b)$ with mean $\mu$ and a scaling paramter $b$ is defined as
$$f_X(x;\mu, b) = \frac{1}{2b} \exp \left( -\frac{|x-\mu|}{b} \right)$$
The standard Laplace distribution is a simplifed version where $\mu = 0$ and $b=1$.
relation with exponential distribution: As shown at Laplace distribution article on Wikipedia, the Laplace random variable $Z\sim Lap(0, 1/\lambda)$ is obtained by the difference $Z = X-Y$ of two iid exponential random variable $X, Y\sim Exp(\lambda)$.
general case: As pointed out in Proposition 2.2.3 from The Laplace distribution and generalizations: a revisit with applications to Communications, Economics, Engineering and Finance by Kotz et al, a Laplace random variable $Z$ is obtained by
$$Z=\sum w_i X_i$$
where each $w_i$ takes on values $\pm 1$ with probabilities 1/2 and $X_i$ is standard exponential random variables.
question: I'm interested in the generalization of this representation so that
$$Z = \sum w_i X_i$$
where each $w_i$ is any real number and each $X_i$ is independently sampled from $Exp(1)$. If $w_i >0$, then it defines the hypoexponential distribution since the the summation becomes $Z=\sum Exp(1/w_i)$ due to the closure under scaling.
Is there any probability distribution about a weighted sum where each $w_i$ is any real number?
 A: In other words, gives two independent random variables $X$ and $Y$, distributed according to hypoexponential distribution with parameters $\{ w_1, w_2, \ldots, w_n \}$ and $\{v_1,v_2, \ldots, v_m \}$ respectively, you are asking to determine the distribution of $Z=X-Y$.
Let $\Theta_X$, and $\Theta_Y$ denote matrices from the probability density functions of respective hypoexponential distributions, see wiki page:
$$
   f_X(x) = - \langle\vec{\alpha}_n, \exp(x \Theta_X) \Theta_X, \vec{1}_n \rangle \cdot [ x > 0 ] \qquad \qquad
   f_Y(y) = - \langle\vec{\alpha}_m, \exp(y \Theta_Y) \Theta_Y, \vec{1}_m \rangle \cdot [ y > 0 ]
$$
where $(\alpha_n)_i = \delta_{i,1}$, $ (\vec{1}_n)_i = 1$ and $(\alpha_m)_j = \delta_{j,1}$, $ (\vec{1}_m)_j = 1$, $i=1,\ldots,n$, and $j=1,\ldots,m$. 
Then 
$$ \begin{eqnarray}
   f_Z(z) &=& \int_{-\infty}^\infty f_X(z+y) f_Y(y) \mathrm{d} y = 
            \int_{\max(-z,0)}^\infty f_X(z+y) f_Y(y) \mathrm{d} y \\ 
          &=& 
            \int_{\max(-z,0)}^\infty \langle \vec{\alpha}_n \otimes \vec{\alpha}_m, \left( \mathrm{e}^{(z+y) \Theta_X} \Theta_X \right) \otimes \left( \mathrm{e}^{y \Theta_Y} \Theta_Y\right), \vec{\mathbf{1}}_n \otimes \vec{\mathbf{1}}_m \rangle   \mathrm{d} y \\
          &=&
\int_{0}^\infty \langle \vec{\alpha}_n \otimes \vec{\alpha}_m, \left( \mathrm{e}^{(\max(z,0)+y) \Theta_X} \Theta_X \right) \otimes \left( \mathrm{e}^{(\max(-z,0) + y) \Theta_Y} \Theta_Y\right), \vec{\mathbf{1}}_n \otimes \vec{\mathbf{1}}_m \rangle   \mathrm{d} y \\ 
&=&
\langle \vec{\alpha}_n \otimes \vec{\alpha}_m, \left( \mathrm{e}^{\max(z,0) \Theta_X} \otimes \mathrm{e}^{(\max(-z,0) ) \Theta_Y} \right) \cdot \left( \int_{0}^\infty \mathrm{e}^{y \Theta_X} \Theta_X \otimes \mathrm{e}^{y \Theta_Y} \Theta_Y \mathrm{d} y  \right), \vec{\mathbf{1}}_n \otimes \vec{\mathbf{1}}_m \rangle
  \end{eqnarray}
$$
The formula above tells the density function for $Z$ will be piecewise, much like Laplace distribution, with functional form of $X$ variate for $z>0$ and functional form of $Y$ variate for $z<0$.
Example:
Consider an example with $n=2$ and $m=2$, and $\{w_1,w_2\} = \{1,2\}$, and $\{v_1,v_2\} = \{1,1\}$. Corresponding matrices are
$$
    \Theta_X = \left( \begin{array}{cc} -1 & 1 \\ 0 & -2 \end{array} \right) \qquad
    \Theta_Y = \left( \begin{array}{cc} -1 & 1 \\ 0 & -1 \end{array} \right)
$$
Then
$$
   \exp\left(x \Theta_X \right) = \left(
\begin{array}{cc}
 e^{-x} & e^{-x}-e^{-2 x} \\
 0 & e^{-2 x} \\
\end{array}
\right) \qquad
   \exp\left(y \Theta_Y \right) = \left(
\begin{array}{cc}
 e^{-y} & e^{-y} y \\
 0 & e^{-y} \\
\end{array}
\right)
$$
$$
   \exp\left(x \Theta_X \right) \Theta_X =
\left(
\begin{array}{cc}
 -e^{-x} & 2 e^{-2 x}-e^{-x} \\
 0 & -2 e^{-2 x} \\
\end{array}
\right)
 \qquad
   \exp\left(y \Theta_Y \right) \Theta_Y = 
\left(
\begin{array}{cc}
 -e^{-y} & e^{-y}-e^{-y} y \\
 0 & -e^{-y} \\
\end{array}
\right)
$$
Using Kronecker product, 
$$
  \int_{0}^\infty \mathrm{e}^{y \Theta_X} \Theta_X \otimes \mathrm{e}^{y \Theta_Y} \Theta_Y \mathrm{d} y = \left(
\begin{array}{cccc}
 \frac{1}{2} & -\frac{1}{4} & -\frac{1}{6} & \frac{7}{36} \\
 0 & \frac{1}{2} & 0 & -\frac{1}{6} \\
 0 & 0 & \frac{2}{3} & -\frac{4}{9} \\
 0 & 0 & 0 & \frac{2}{3} \\
\end{array}
\right)
$$
Combining things, with little algebra we get:
$$
  f_Z(z) =  \left\{
\begin{array}{cc}
 \frac{5}{18} & z=0 \\
 \frac{1}{18} \mathrm{e}^{-z} \left(9-4 \mathrm{e}^{-z}\right) & z>0 \\
 \frac{1}{18} \mathrm{e}^z (5-6 z) & z < 0 \\
\end{array} \right.
$$
A: 
a Laplace random variable $Z$ is obtained by
  $$Z=\sum w_iX_i$$
  where each $w_i$ takes on values $\pm 1$ with probabilities $1/2$ and $X_i$ is standard exponential random variables.

Is this statement really correct?  I can see why it holds when $Z = w_1X_1$ because
the conditional density of $Z$ given $w_1 = +1$ is $\exp(-x)\mathbf 1_{[0,\infty)}$ while the conditional density of $Z$ given $w_1 = -1$ is $\exp(x)\mathbf 1_{(-\infty,0]}$, and so by the law of total probability, we have
$$f_Z(x) = \frac{1}{2}\exp(-x)\mathbf 1_{[0,\infty)}
+ \frac{1}{2}\exp(x)\mathbf 1_{(-\infty,0]} = \frac{1}{2}\exp(-|x|), 
-\infty < x < \infty,$$
which, ignoring the value at $x=0$, is a Laplacian density.  But by the 
same argument, the conditional density of $w_1X_1 + w_2X_2$ is a Gamma
density of order $2$ when $w_1 = w_2 = +1$, and this should show up in
the unconditional pdf as well.
Turning to the OP's question about the density of $\sum w_iX_i$ where
the $w_i$ are arbitrary real numbers and the $X_i$ are independent 
exponential random variables with mean $1$, note that the moment-generating
function of $w_iX_i$ is $E[\exp(tw_iX_i)] = (1-w_it)^{-1}$, we have
$$E[\exp(tZ)] = \prod_{i=1}^n \frac{1}{1 - w_it}.$$
Assuming that the $|w_i|$ all are distinct real numbers, this can be expanded
via partial fractions into a weighted sum of terms of the 
form $(1-w_it)^{-1}$, and so the density of $Z$ is a weighted sum of
exponential densities $w_i^{-1}\exp(-x/w_i)\mathbf 1_{[0,\infty)}$ (when
$w_i > 0$) and densities $|w_i|^{-1}\exp(-x/w_i)\mathbf 1_{(-\infty,0]}$ (when
$w_i < 0$).  
