$u_x + u^2u_y = 1$ with $u(x,0)=1$

How can we solve $u_x + u^2u_y = 1$ with $u(x,0)=1$? I have reached a dead end. I'm using the characteristic relations $$\frac{dx}{dt}=1,\enspace \frac{dy}{dt}=u^2,\enspace \frac{du}{dt}=1$$

The first implies that $dx=dt$. We can use this in the third relation and see that $$\frac{du}{dx}=1$$

which would mean that $u=x+f(y)$. Why is this wrong?

You set up the characteristic equations correctly, but you're forgetting the variables $x,y,u$ in your characteristic equations are coupled. The equation $\frac{du}{dx}=1$ is not the same as $u_x=1.$ Here is the standard procedure for solving a DE of this sort using the method of characteristics.

Solve for $x$ and $u$ to get $x=t+C_1$ and $u=t+C_3$. This means $\frac{dy}{dt}=(t+C_3)^2$ and so $y=\frac{1}{3}(t+C_3)^3+C_2$. Enforcing the condition that $(x,y,z)=(s,0,1)$ whenever $t=0$ ensures that our surface contains the line $y=0,z=1$ and satisfies the initial condition $u(x,0)=1$. This yields $C_1=s,C_2=-\frac{1}{3},C_3=1$ and provides us with a parametrization of our solution $$x=t+s \\ y=\frac{1}{3}(t+1)^3-\frac{1}{3} \\ u=t+1$$ Hence $$y=\frac{u^3}{3}-\frac{1}{3} \iff u(x,y)=(3y+1)^{1/3}$$