Only a partial answer : (2) seems strange. In dimension $1$, if $u$ does not change sign, your setting includes the one of $u^\alpha \in H^1(0,1)$ (boundary values are irrelevant here) for some $\alpha>1$ and you're asking if it's enough to get $u' \in L^{2\alpha}(0,1)$.

For $\alpha>1$ and $u(x)=x^{\frac{1}{\alpha}}$ then $u^\alpha$ certainly belongs to $H^1(0,1)$ but the singularity of $u'(x) = \frac{1}{\alpha} \frac{1}{x^{1-\frac{1}{\alpha}}}$ is not in $L^{2\alpha}(0,1)$, at least for $\alpha$ large enough.

EDIT:

I think (1) implies, one way or another, that $D(u)$ is locally integrable, at least I don't understand this equality if it's not the case ; the counterexample below shows that this is false in general so for (1) to be satisfied, I believe something's missing.

The function $u:x \mapsto x\cos(1/x)$ is bounded on $L^\infty(0,1)$ but is not in $W^{1,1}(0,1)$. Indeed, $$u'(x)=\cos(1/x)+\frac{\sin(1/x)}{x},$$

where the first part of the sum is bounded and the second not integrable on $(0,1)$ because related to the improper integral of $z\mapsto\frac{\sin(z)}{z}$ on $(1,+\infty)$. However, computing the derivative of $u^3=|u|^2u$ we get
$$
3u(x)^2u'(x) = 3 x^2\cos(1/x)^3 + 3x\cos(1/x)^2\sin(1/x),
$$
which is bounded on $(0,1)$. We have therefore $|u|^2 u\in H^1(0,1)$ but $u\notin W^{1,1}(0,1)$.

Now, as noticed by the Perelman in the comments, this is not yet a counterxample because $u$ is not $W^{1,1}(0,1)$ but yet is $W^{1,1}_{\text{loc}}(0,1)$. However, replacing $u$ by its reflection $v\in L^\infty(-1,1)$ which is even and equals $u$ on $(0,1)$, I think $v'$ is not $W^{1,1}_{\text{loc}}(-1,1)$.

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