Where do we need absolute values when solving $u'(t) = \frac{u^2(t) - u(t)}{t}$

Hint.

$$ (t u)' = u^2 $$

now making $ v = t u$

$$ v' = \frac{v^2}{t^2}\Rightarrow \frac{dv}{v^2} = \frac{dt}{t^2} $$

or integrating

$$ \frac 1v = \frac 1t + C\Rightarrow v = \frac{t}{C t+1} = u t $$

The maximal $(t,u)$-domain relevant to the given IVP is $\Omega:={\mathbb R}_{>0}\times{\mathbb R}$. Within $\Omega$ the standard existence and uniqueness theorem for ODEs is valid. By inspection one sees that there are the constant solutions $u_0(t)\equiv0$ and $u_1(t)\equiv1$ $(t>0)$. No other solution can cross the graphs of $u_0$ and $u_1$. Since the initial point $\bigl(1,{1\over2}\bigr)$ is given the solution $t\mapsto u(t)$ therefore stays in the $u$-interval $\>]0,1[\>$ for all $t>0$. Hence there are no case distinctions; you just have to select the primitive of $s\mapsto{1\over s^2-s}$ that is valid for $0<s<1$, namely $$\eqalign{\int{ds\over s^2-s}&=\int\left({1\over s-1}-{1\over s}\right)ds=\log|s-1|-\log |s| +C\cr &=\log{1-s\over s}+C\qquad(0<s<1)\ .\cr}$$