Solution attempt $xu_x+u_y-(y+z)u_z=0$ Solve $$
\begin{cases}
xu_x+u_y-(y+z)u_z=0\\
u(x,1,z)=xz\\
\end{cases}
$$
I got:
\begin{cases}
x_t=x\\
y_t=1\\
z_t=-y-z\\
u_t=0
\end{cases}
Therefore:
\begin{cases}
x(t,s_1,s_2)=e^tf_1(s_1,s_2)\\
y(t,s_1,s_2)=t+f_2(s_1,s_2)\\
z(t,s_1,s_2)=e^{-t}f_3(s_1,s_2)-t-f_2(s_1,s_2)\\
u(t,s_1,s_2)=f_4(s_1,s_2)
\end{cases}
Using initial condition:
\begin{cases}
s_1=x=f_1(s_1,s_2)\\
1=y=f_2(s_1,s_2)\\
s_2=f_3(s_1,s_2)-f_2(s_1,s_2)\\
s_1s_2=u=f_4(s_1,s_2)
\end{cases}
So:
\begin{cases}
x(t,s_1,s_2)=e^ts_1\\
y(t,s_1,s_2)=t+1\\
z(t,s_1,s_2)=e^{-t}(s_2+1)-t-1\\
u(t,s_1,s_2)=s_1s_2
\end{cases}
I am writing $$u=s_1s_2$$ by $x,y,z$ term and get $$x(z+y)-xe^{1-y}$$
which is incorrect, where is the problem?
 A: $$xu_x+u_y-(y+z)u_z=0$$
The Charpit-Lagrange system of characteristic ODEs is :
$$\frac{dx}{x}=\frac{dy}{1}=\frac{dz}{-(y+z)}$$
A first characteristic equation comes from $\frac{dx}{x}=\frac{dy}{1}$
$$\frac{1}{x}e^y=c_1$$
A second characteristic equation comes from $\frac{dy}{1}=\frac{dz}{-(y+z)}$
The solution of this ODE is :$\quad z=c_2e^{-y}-y+1$ 
$$(z+y-1)e^y=c_2$$
The general solution of the PDE on the form of implicit equation $u=F(c_1\:,\;c_2)$ is :
$$\boxed{u(x,y,z)=F\big(\frac{1}{x}e^y,(z+y-1)e^y\big)}$$
The arbitrary function $F$ has to be determined according to the specified condition.
CONDITION : $u(x,1,z)=xz$
$$xz=F\big(\frac{1}{x}e,(z+1-1)e\big)$$
Let $X=\frac{e}{x}\quad\implies\quad x=\frac{e}{X}$
and $Y=ez\quad\implies\quad z=\frac{Y}{e}$
$xz=\frac{e}{X}\frac{Y}{e}=\frac{Y}{X}=F(X,Y)$
Now the function $F$ is determined :
$$F(X,Y)=\frac{Y}{X}$$
We put it into the above general solution where $X=\frac{1}{x}e^y$ and $Y=(z+y-1)e^y$
$$u(x,y,z)=\frac{(z+y-1)e^y}{\frac{1}{x}e^y}$$
$$\boxed{u(x,y,z)=x(z+y-1)}$$
$$ $$
ADDITION as an answer to OP's comment.
The OP asked why $\frac{1}{x}e^y=c_1$ was used instead of $xe^{-y}=C_1$.
Answer : Both can be used equivalently. For example with the second $xz=\Phi(C_1,c_2)$ 
$$xz=\Phi\big(xe^{-1},(z+1-1)e\big)$$
Let $X=xe^{-1}\quad\implies\quad x=eX$
and $Y=ez\quad\implies\quad z=\frac{Y}{e}$
$xz=eX\frac{Y}{e}=XY=\Phi(X,Y)$
Now the function $\Phi$ is determined :
$$\Phi(X,Y)=XY$$
We put it into the above general solution where $X=xe^{-y}$ and $Y=(z+y-1)e^y$
$$u(x,y,z)=(xe^{-y})(z+y-1)e^y$$
$$\boxed{u(x,y,z)=x(z+y-1)}$$
$$ $$
COMMENT :
The system of equations that you wrote :
$\begin{cases}
\frac{dx}{dt}=x\\
\frac{dy}{dt}=1\\
\frac{dz}{dt}=-y-z\\
\frac{du}{dt}=0
\end{cases}$
is exactly he same as the Charpit-Lagrange written on this form 
$$dt=\frac{dx}{x}=\frac{dy}{1}=\frac{dz}{-z-y}=\frac{du}{0} .$$
Of course, both lead to the same characteristic equations. Fundamentally there is no difference except presentation and symbolism. The approach that you use which introduces a lot of new parameters is especially usefull when it is simpler to express the result on the form of parametric solution. In the present case where the result can be expressed on explicit form, one can discuss the interest of introducing more parameters which are eliminated at the end. But it doesn't matter, that's up to you to chose the approach appearing the most intelligible for you. 
It seems that there is a mistake in your calculus when solving the coupled equations $\frac{dy}{dt}=1\quad,\quad\frac{dz}{dt}=-y-z$.
Unfortunately the intermediate steps are not edited and the further steps not sufficiently detailed. One cannot check them. Possibly you should get :
\begin{cases}
x(t,s_1,s_2)=e^tf_1(s_1,s_2)\\
y(t,s_1,s_2)=t+f_2(s_1,s_2)\\
z(t,s_1,s_2)=e^{-t}f_3(s_1,s_2)-t-f_2(s_1,s_2)+1\\
u(t,s_1,s_2)=f_4(s_1,s_2)
\end{cases}
$+1$ is missing in your third equation. As a consequence your further calculus has to be slightly corrected.
Using initial condition:
\begin{cases}
s_1=x=f_1(s_1,s_2)\\
1=y=f_2(s_1,s_2)\\
s_2=f_3(s_1,s_2)-f_2(s_1,s_2)+1\\
s_1s_2=u=f_4(s_1,s_2)
\end{cases}
So:
\begin{cases}
x(t,s_1,s_2)=e^ts_1\\
y(t,s_1,s_2)=t+1\\
z(t,s_1,s_2)=e^{-t}s_2-t\\
u(t,s_1,s_2)=s_1s_2
\end{cases}
and finally
$s_1=e^{-t}x$
$s_2=e^t(z+y-1)$
$$u=s_1s_2=x(z+y-1)$$
