Applying method of characteristic equations to $e^y u_x + u_y = u^2$ I have the PDE  $e^y u_x + u_y = u^2$, $u(x,0) = x$ for small $|y|$. Also $ u = u(x(s), y(s))$.
So $\frac{dx}{ds} = e^{y}$, $\frac{dy}{ds} = 1$, and $\frac{du}{ds} = u^{2}$.
Solving the first ode: $\frac{dx}{ds} = e^{y} \to dx = e^{y}ds$ and integrate both sides to obtain $x = e^{y}s + c_1$.
The second ODE: $\frac{dy}{ds} = 1 \to dy = 1ds$ and integrate both sides to obtain $y = s + c_2$.
The third ODE: $\frac{du}{ds} = u^{2}$, the solution to this ode is $-\frac{1}{u} = s + c_3 \implies u = -\frac{1}{s+c_3}$. 
Okay now I want to eliminate the parameter $s$. I solve for $s$ from the solution of the second ode: $s = y -c_2$ then substitute into the solution of the first ode: $x = e^{y}(y -c_2) + c_1 = e^{y}y - e^{y}c_2 + c_1$. Also $u = -\frac{1}{y-c_2 + c_3}$.
This is where I am stuck and am not sure what to do. I was attempting to follow https://en.wikipedia.org/wiki/Method_of_characteristics#Example but I am not doing this correctly I think.
 A: $$e^yu_x+u_y=u^2$$
I agree with your equations which can be written in a  summarised form :
$$\frac{dx}{e^y}=\frac{dy}{1}=\frac{du}{u^2}=ds$$
A first characteristic equation comes from $\quad\frac{dx}{e^y}=\frac{dy}{1}\quad$ leading to :
$$e^y-x=c_1$$
A second characteristic equation comes from $\quad\frac{dy}{1}=\frac{du}{u^2}\quad$ leading to :
$$\frac{1}{u}+y=c_2$$
The general solution of the PDE, expressed on the form of implicit equation is :
$$\Phi\left((e^y-x) \:,\: (\frac{1}{u}+y) \right)=0$$
$\Phi$is an arbitrary function of two variables.
Or, equivalently the general solution of the PDE on explicit form is :
$$\frac{1}{u}+y=F(e^y-x)$$
where $F$ is an arbitrary function.
$$u(x,y)=\frac{1}{-y+F(e^y-x)}$$
CONDITION :
$u(x,0)=x=\frac{1}{-0+F(e^0-x)}=\frac{1}{F(1-x)}$
$F(1-x)=\frac{1}{x}$
Let $\quad X=1-x\quad;\quad x=1-X\quad;\quad F(X)=\frac{1}{1-X}$ 
So, the function $F$ is determined. 
We put it into the above general solution where $\quad X=e^y-x\quad$ then $\quad F(e^y-x)=\frac{1}{1-(e^y-x)}$
$$u(x,y)=\frac{1}{-y+\frac{1}{1-e^y+x}}$$
This is the particular solution of the PDE which fits to the boundary condition.
Note: The above calculus is consistent with your calculus. The difference is that $ds$ is eliminated at the beginning,  while in your calculus $ds$ is eliminated later. This doesn't change the final result.
ADDITION, answering to the comment of MasterYoda :
Of course, I checked the solution $u(x,y)=\frac{1}{-y+\frac{1}{1-e^y+x}}$ before publishing my answer. After the MasterYoda's comment I checked it again. Definitively the solution satisfies the PDE (copy below).

