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