I am attempting to solve the differential equation: $$(y')^2 = 1 $$ with the initial condition $y(0)=0$
By moving into the Laplace domain we get:
\begin{align*}s^{2}Y(s)^{2} &= \frac{1}{s}\\Y(s)^{2} &= \frac{1}{s^3}\\Y(s) &= \pm\frac{1}{s^{3/2}}\\ \left(\frac{1}{2}\right)!Y(s) &= \pm\frac{\left(\frac{1}{2}\right)!}{s^{3/2}}\\\left(\frac{\sqrt{\pi}}{2}\right)\cdot Y(s) &=\pm\frac{\left(\frac{1}{2}\right)!}{s^{3/2}}.\end{align*}
which gives 2 solutions:
\begin{align*}Y(s) &= \frac{2\left(\frac{1}{2}\right)!}{\sqrt{\pi}s^{3/2}},\\Y(s) &= -\frac{2\left(\frac{1}{2}\right)!}{\sqrt{\pi}s^{3/2}}.\end{align*}
Now returning to the time domain,
$$y(t) =\pm \frac{2}{\sqrt{\pi}} t^{1/2}.$$
Did I solve it correctly?
As others said in the comments of your post, the Laplace transform does not commute with non linear operations such as squaring the function.
The equation can be rewritten as $y' = \pm 1$ which has many weak solutions (the $\pm$ depends on $t$):
If $A\subset\mathbb{R}$ is a measurable set and $\lambda$ is the Lebesgue measure, the continuous function \begin{equation} y_A(t)=2\,\text{sgn}(t)\lambda\left(A\cap (0,t)\right)-t \end{equation} satisfies $y_A(0) = 0$ and in the sense of the distributions \begin{equation} y_A'= \mathbf{1}_A - \mathbf{1}_{A^c} \end{equation} Hence $(y_A')^2 = 1$ a.e.