I don't know how your prove can work without prove the trace estimation...
Here is a more standard idea. (You can find it on Evans PDE book or Leoni's Sobolev space. I would say Evans, it is easier)
Suppose $\Omega$ is an bounded extension domain so that the extension theorem, which you quote in your post, will work. Next, for $u\in W^{1,p}(\Omega)$ you actually can obtain a sequence $(u_n)\subset C^\infty(\bar\Omega)\cap W^{1,p}(\Omega)$ such that $u_n\to u$ in $W^{1,p}$ norm. This is an standard result of global approximation and you could prove it easily by using extension theorem.
Next, you could prove that, and this is an very important result in trace theorem (I would say the most important), the Trace estimation
$$ \|T [u_n]\|_{L^p(\partial \Omega)}\leq C \|u_n\|_{W^{1,p}(\Omega)}\tag 1$$
Remember here $u_n\in C^\infty(\bar\Omega)$, hence $T[u_n]=u\lfloor_{\partial\Omega}$ is well defined. (This estimation also )
Hence, you have
$$ \|T [u_n]-T[u_m]\|_{L^p(\partial \Omega)}\leq C \|u_n-u_m\|_{W^{1,p}(\Omega)}\tag 2 \to 0$$
which implies that $T[u_n]$ is a cauchy sequence in $L^p(\partial\Omega)$ and hence you could define the limit and this limit is $T[u]$.
Finally, you have
$$\|T [u_n]-T[u]\|_{L^p(\partial \Omega)}\leq C \|u_n-u\|_{W^{1,p}(\Omega)}\to 0$$
which shows $u_n\to u$ a.e. on $\partial \Omega$, which addressed your question.
I will sketch two possible ways; tell me if you need more details at some point.
Choose a finite cover of $\partial\Omega$ made of open sets
$U_i\Subset V$ for which a $C^1$-diffeomorphism $\psi_i:U_i\to B^{n-1}\times (-1,1)$ exists. Here we are thinking a point $x\in\mathbb{R}^n$ as a couple
$(x',t)\in\mathbb{R}^{n-1}\times\mathbb{R}$, so the last set is
$$B^{n-1}\times (-1,1)=\{x=(x',t):|x'|<1,\ |t|<1\}.$$
The diffeomorphism $\psi_i$ is required to "straighten the boundary",
i.e. $\psi_i(U_i\cap\partial\Omega)=B^{n-1}\times\{0\}$; we also require that
$\psi_i(U_i\cap\Omega)=B^{n-1}\times(0,1)$. In words, we are mapping $U_i\cap\Omega$ to the upper half of the cylinder $B^{n-1}\times(-1,1)$.
(Why do such $U_i$'s and $\psi_i$'s exist?)
Using partitions of unity $\phi_i$ (subordinate to the cover $\{U_0\}\cup\{U_i\}$ of $\overline{\Omega}$, where $U_0$ covers the remaining part of the interior, i.e. $\Omega\setminus\cup U_i\subset U_0\Subset\Omega$) and looking at the functions $v_i:=(\phi_i u)\circ\psi_i^{-1}$, we reduce to the following problem: given $v\in W^{1,\infty}(B^{n-1}\times(0,1))$, we have to build an extension $Ev\in W^{1,\infty}(B^{n-1}\times(-1,1))$ in such a way that $E$ is a linear bounded operator and maps functions such as our $v_i$'s to some function lying also in $W^{1,\infty}(\mathbb{R}^n)$.
Then it will suffice to sum all the functions $Ev_i\circ\psi_i$, together with $\phi_0u$, to obtain the desired extension.
This is simply done by reflection:
put $Ev(x',t):=v(x',t)$ if $t>0$ and $Ev(x',t):=v(x',-t)$ if $t<0$.
It is easy to see that $\frac{\partial (Ev)}{\partial x_i}$ exists and is given by the same formula for $i<n$: to check the definition of weak derivative against a $\phi\in C^1_c(B^{n-1}\times(-1,1))$ one can first consider $\phi\cdot\eta_k$, where $\eta\in C^\infty(\mathbb{R})$ satisfies $\eta(t)=0$ for $t<\frac{1}{2}$, $\eta(t)=1$ for $t>1$, and $\eta_k(t):=\eta(k t)$; then one lets $k\to\infty$.
We claim that $\frac{\partial (Ev)}{\partial x_n}$ also exists and equals $g$, where $g(x',t)=\frac{\partial v}{\partial x_n}(x',t)$ for $t>0$, while $g(x',t)=-\frac{\partial v}{\partial x_n}(x',t)$ for $t<0$. To see this, pick any test function $\phi\in C^1_c(B^{n-1}\times(-1,1))$ and observe that
$$\int_{B^{n-1}\times (-1,1)}Ev\frac{\partial \phi}{\partial x_n}=\int_{B^{n-1}\times (0,1)}v\frac{\partial\chi}{\partial x_n}$$
where $\chi(x',t):=\phi(x',t)-\phi(x',-t)$. Since $\chi(x',0)=0$, we have the estimate $|\chi(x',t)|\le Mt$ for $t>0$, where
$M:=\|\frac{\partial\chi}{\partial x_n}\|_\infty$. Now cutoff $\chi$ using $\eta_k$ as before, and apply the definition of weak derivative:
$$\int_{B^{n-1}\times (0,1)}\frac{\partial v}{\partial x_n}(\eta_k\chi)=-\int_{B^{n-1}\times (0,1)}v\frac{\partial(\eta_k\chi)}{\partial x_n}$$
$$=-\int_{B^{n-1}\times (0,1)}k\eta'(kt)v(x',t)\chi(x',t)-\int_{B^{n-1}\times (0,1)}\eta_k v\frac{\partial\chi}{\partial x_n}$$
(here we used sometimes $x_n$ and sometimes $t$ to denote the last component of $x$).
To conclude, it suffices to prove that $\int_{B^{n-1}\times (0,1)}k\eta'(kt)v(x',t)\chi(x',t)\to 0$ (as $k\to\infty)$: then in the limit we deduce
$$\int_{B^{n-1}\times (0,1)}\frac{\partial v}{\partial x_n}\chi=-\int_{B^{n-1}\times (0,1)}v\frac{\partial\chi}{\partial x_n}$$
and the left hand side equals $\int_{B^{n-1}\times(-1,1)}g\phi$,
while the right hand side is $-\int_{B^{n-1}\times(-1,1)}Ev\frac{\partial \phi}{\partial x_n}$, so we are done. But
$$\left|\int_{B^{n-1}\times (0,1)}k\eta'(kt)v(x',t)\chi(x',t)\right|\le kCM\int_{B^{n-1}\times(0,\frac{1}{k})}|v|t
\le CM\int_{B^{n-1}\times(0,\frac{1}{k})}|v|\to 0,$$
where we put $C:=\|\eta'\|_\infty$.
This extension operator has the defect of mapping $C^1(\overline{\Omega})$ outside $C^1(V)$. To circumvent this, another more clever construction is as follows:
let's take for granted the extension theorem for $W^{1,1}$ (which comes from what we have done up to now).
Put $Ev(x',t):=v(x',t)$ if $t>0$ and $Ev(x',t):=-3v(x',-t)+4v(x',-\frac{t}{2})$ if $t<0$. The key point now is that it suffices to show that $Ev\in W^{1,1}$ because then (restricting to the two halves) the weak gradient of $Ev$ is forced to be in $L^\infty$ (rather than only in $L^1$) as we have an explicit formula for it (write it down!).
For any $W=B^{n-1}_{1-\epsilon}\times(-1+\epsilon,1-\epsilon)\Subset B^{n-1}\times (-1,1)$ we can find a sequence $(v_n)\subset W^{1,1}(W)\cap C^1$ converging to $v$ in $B^{n-1}_{1-\epsilon}\times(0,1-\epsilon)$
(use the extension for $W^{1,1}$ and mollify). Then $Ev_n\in C^1$ (check it),
where of course we are now restricting to $W$, and
$$\|Ev_n-Ev_m\|_{W^{1,1}}=\|E(v_n-v_m)\|_{W^{1,1}}\le C\|v_n-v_m\|_{W^{1,1}}.$$
Thus $(Ev_n)\subset W^{1,1}(W)$ is a Cauchy sequence. But it has to converge to $v$, so (at least locally) $v\in W^{1,1}$.
Best Answer
Expanding the comment by user127096.
The space $W^{1,\infty}({\cal{O}})$ is equal to the space of bounded locally Lipschitz functions. This means for every element $u\in W^{1,\infty}({\cal{O}})$ it is possible to find a function $v$ that is Lipschitz continuous by just rearranging a set of measure zero.
The trace then can be taken as you take the trace of continuous functions by pointwise convergence.