An elementary proof was shown to me by Jan Malý. It has never been published and with his permission we included it in [1] (Proposition 28). The argument is very elementary (1.5 pages with all details), but still too long to be included here.
Theorem. For $n\geq 1$ and $p>1$, there is a bounded linear extension operator $$ E:W^{1,p}(\mathbb{R}^{n})\to W^{1,q}\cap C^\infty(\mathbb{R}^{n+1}_+), \quad \text{where $q=\frac{(n+1)p}{n}$.} $$ In other words, $W^{1,p}(\mathbb{R}^{n})$ continuously embeds into the trace space $W^{1-\frac{1}{q},q}(\mathbb{R}^{n})$ of $W^{1,q}(\mathbb{R}^{n+1}_+)$.
From this result the following corollary follows right away:
Corollary. If $\Omega\subset\mathbb{R}^n$, $n\geq 2$, is a bounded and smooth domain, then there is a bounded extension operator $$ E:W^{1,p}(\partial\Omega)\to W^{1,q}\cap C^\infty(\Omega), \quad \text{where $1<p<\infty$ and $q=\frac{np}{n-1}$.} $$
Taking $p=n-1$ and $n>2$ yields the theorem asked in the question. We need to take $n>2$ since for $n=2$ we have $p=n-1=1$ and the result is false in that case (there are counterexamples, see [1]).
[1] P. Goldstein, P. Hajłasz, Jacobians of $W^{1,p}$ homeomorphisms, case $p=[n/2]$. Calc. Var. Partial Differential Equations 58 (2019), no. 4, Art. 122, 28 pp.