The quasilinear first order PDE $ a\big(x,y,u(x,y)\big) u_x(x,y) + b\big(x,y,u(x,y)\big)u_y(x,y) = c\big(x,y,u(x,y)\big) $ where $a,\,b,\,c \in C^1$ with data $\mathcal{C}(\xi) = \big(x(\xi), y(\xi), u(\xi)\big) \in C^1$ and with $ \begin{vmatrix} \frac{dx}{d\xi} & a \\ \frac{dy}{d\xi} & b\end{vmatrix} \neq 0 $ has a unique solution near $\mathcal{C}$ given by \begin{align} \frac{d x}{d \eta} &= a & x\big|_{\eta = 0}&= x(\xi)\\ \frac{d y}{d \eta} &= b & y\big|_{\eta = 0}&= y(\xi)\\ \frac{d u}{d \eta} &= c & u\big|_{\eta = 0}&= u(\xi)\\ \end{align}
For proof and geometrical interpretation, see F. John's Partial Differential Equations (§1.4)
In your case, $\mathcal{C}(\xi) = \big(\xi,0,\xi+2\big)$. Near $\eta \sim 0$ $ \begin{vmatrix} \frac{dx}{d\xi} & a \\ \frac{dy}{d\xi} & b\end{vmatrix} = \begin{vmatrix} 1 & u^2 \\ 0 & 1\end{vmatrix} = 1 $ and the solution is unique.
The system of ODE's is \begin{align} \frac{d x}{d \eta} &= u^2 & x\big|_{\eta = 0}&= \xi\\ \frac{d t}{d \eta} &= 1 & t\big|_{\eta = 0}&= 0\\ \frac{d u}{d \eta} &= 0 & u\big|_{\eta = 0}&= \xi + 2\\ \end{align} with solution $ t = \eta, \quad u = \xi + 2, \quad x = (\xi + 2)^2 \eta + \xi. $
The characteristics are $t = \frac{x - \xi}{(\xi + 2)^2}$ hence $\xi = -2$ is a special point. As $\xi \rightarrow \infty$, $t \rightarrow 0$. As $\xi \rightarrow -\infty$, $t \rightarrow 0$. As $\xi \rightarrow -2$, $t \rightarrow \infty$.

This of course, means that there is no solution when the characteristics meet.
A simple explanation for this is that the transformation $ (x,t) \rightarrow (\xi,\eta) $ is invertible iff $ \begin{vmatrix} \partial_\xi x & \partial_\eta x \\ \partial_\xi t & \partial_\eta t \end{vmatrix} = 1 + 4\eta + 2\xi \eta \neq 0 $ meaning there is no solution when $\xi = -\frac{1 + 4 \eta}{2\eta}$ or, inverting the transformation, when $ t = - \frac{1}{4(x+2)} $
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Lastly, inverting for $\xi$
$ \xi = \frac{-(1 + 4t) \pm \sqrt{1 + 4t(2 + x)}}{2 t} $
and
$ u(x,t) = \frac{-1 \pm \sqrt{1 + 4t(2 + x)}}{2 t}. $
In order to determine the correct sign, we must look at the initial condition. For the minus sign $\lim_{t \rightarrow 0} u(x,t) = -\infty$, while the plus sign gives the correct answer. Hence
$ u(x,t) = \frac{-1 + \sqrt{1 + 4t(2 + x)}}{2 t}. $