A proof for $\nabla |u|^p = p\ \text{sgn}(u)|u|^{p-1}\nabla u$.

An alternative approach is to truncate your function $F(t) = |t|^p.$ For each $M>0$ define $T_M(x) = \min\{|t|,M\}$ and consider $t \mapsto T_M(t)^p.$ This is a Lipschitz function that is piecewise $C^1,$ so the chain rule for Sobolev functions applies ($\dagger$).

Therefore if $u \in W^{1,p}(\Omega),$ we have $T_M(u)^p \in W^{1,1}(\Omega)$ with, $$ \nabla (T_M(u)^p) = \mathbf 1_{\Omega_M}\,p\, \mathrm{sgn}(u) |u|^{p-1}\nabla u, $$ where $\Omega_M = \{ x \in \Omega : |u(x)| \leq M \}.$ Note that $T_M(u)^p \in W^{1,p}(\Omega)$ also, but we will only be able to show that the limit $|u|^p$ lies in $W^{1,1}(\Omega).$

There's a couple of ways to justify the limit as $M \rightarrow \infty,$ and the convergence result I like to use is the following:

Lemma: If $f_n, f \in L^1(\Omega)$ such that $f_n \rightarrow f$ almost everywhere and $\int_{\Omega} |f_n| \,\mathrm{d} x \rightarrow \int_{\Omega} |f| \,\mathrm{d} x,$ then $f_n \rightarrow f$ in $L^1(\Omega).$

This can be proven by reproducing the argument for the dominated convergence theorem (see for example here for details).

One you have this, it's easy to see how to proceed. By monotone convergence, as $M \rightarrow \infty$ we get $$ \int_{\Omega_M} \left|\nabla(T_M(u)^p) \right|\,\mathrm{d}x \rightarrow \int_{\Omega} p |u|^{p-1}|\nabla u| \,\mathrm{d}x \leq p \lVert u\rVert_{L^p(\Omega)}^{p-1} \lVert \nabla u \rVert_{L^p(\Omega)} \leq p\lVert u \rVert_{W^{1,p}(\Omega)}^p. $$ Therefore we get $\nabla(T_M(u)^p) \rightarrow p\,\mathrm{sgn}(u)|u|^{p-1}\nabla u$ in $L^1(\Omega).$ Since $T_M(u)^p \rightarrow |u|^p$ in $L^1(\Omega),$ we get convergence in $W^{1,1}(\Omega)$ and hence the result follows.

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($\dagger$) It is generally true that if $F$ is Lipschitz and $u \in W^{1,p}(\Omega),$ then $F(u) \in W^{1,p}(\Omega),$ (added later) but the proof of this is not obvious. They key insight is to show that if $u \in W^{1,1}_{\mathrm{loc}}(\Omega)$ and $A \subset \mathbb R$ is a null set, then $\nabla u = 0$ a.e. on $u^{-1}(A).$ This means that $F'(u)\nabla u$ is defined regardless of the representative of $F'$ we choose. A proof of chain rule and the above fact can be found in the following:

J. Serrin, D. Varberg. "A General Chain Rule for Derivatives and the Change of Variables Formula for the Lebesgue Integral." The American Mathematical Monthly, Vol. 76, No. 5 (May, 1969), pp. 514-520.

In our case however, we can simply write $T_M(u)^p = g(T_M(u))$ where $g : \mathbb R \rightarrow \mathbb R$ is a $C^1$ function with bounded derivative such that $g(t) = t^p$ on $[0,M].$ Then the result follows by applying the $C^1$ and absolute value cases.

I think you can use the following trick to solve the issue. Take two sequences of smooth functions $f_{\epsilon} \to f, g_{\epsilon} \to g$ in $L^p$, then

$$ |f_{\epsilon}|^{p-1} g_{\epsilon} \to |f|^{p - 1} g \qquad \text{ in } \ L^1. $$

This follows (by adding and subtracting things) from the estimate: $$ \int \big\vert |f_{\epsilon}|^{p-1} g_{\epsilon} \big\vert \le \|f_{\epsilon}\|_{L^p}^{\frac{p}{q}} \| g_{\epsilon}\|_{L^p} $$ with $1/q +1/p = 1.$ This is Holder's inequality.

Then your result follows by taking a smooth approximation of $u$ and letting $f = u, g = \mathrm{sgn}(u) \nabla u .$

A nice proof in Lieb and Loss's Analysis Chapter 6 (Distributions) page 152.