I do not know a reference to a published proof, most sources refer to Blanchard's paper. The proof is relatively easy once you remember the structure of a canonical graph representation of $\beta$-shifts. Assume that $\beta>1$ is such that the greedy $\beta$-expansion $(\omega_i)_{i\ge 1}$ of $\mathbf 1$ is not eventually periodic (as otherwise the $\beta$-shift is sofic, hence synchronized). The canonical representation of this $\beta$-shift is a countable graph with vertices $v_0,v_1,v_2,\ldots$ and right edges $v_{i-1}\to v_i$ labeled with $\omega_i$ for $i=1,2,\ldots$ and for each vertex $v_{i-1}$ which is an initial vertex of some right edge labeled with nonzero symbol (that is, for each $i\ge 1$ such that $\omega_i>0$) we add backward edges, that is, we add $\omega_i$ different edges (each of them starts at $v_{i-1}$ and ends at $v_0$) labeled by $0,1,\ldots,\omega_i-1$ (that is, there is $\omega_i$ such edges). Then a finite word $w=w_1\ldots w_n$ is in the language of the $\beta$-shift $S_\beta$ if and only if there is a path in our graph starting at $v_0$ and labeled with $w$. Now if the greedy $\beta$-expansion $(\omega_i)_{i\ge 1}$ of $\mathbf 1$ is not dense in $S_\beta$ then there exists a word $u=u_1\ldots u_k$ in the language of $S_\beta$ which does not appear as a label for a path consisting of right edges only. Take a shortest such word. It means that $u_1\ldots u_{k-1}$ is a label for a path consisting of right edges only. Consider a path starting at $v_0$ following right edges until the first appearance of $u_1\ldots u_{k-1}$ and then returning to $v_0$ so that the word $u$ is a suffix of the label of that path. Let $\bar u=\bar u_1\ldots \bar u_\ell$ be the label of that path. There is exactly one backward edge on that path. Note that it follows that $\bar u_1\ldots \bar u_{\ell-1}$ is a ``lexicographically maximal'' label and such a label can only appear over a path consisting of right edges only (this is a defining property of $\beta$-shifts). I claim that $\bar u$ is a synchronizing word. This is because each path labeled by $\bar u$ and appearing somewhere in the graph must end at $v_0$ as otherwise $\bar u$ would appear in the greedy $\beta$-expansion of $\mathbf 1$. Let $w_1\bar u$, $\bar uw_2$ be two words in the language of $S_\beta$. Then $w_1\bar uw_2$ is a label for a path starting at $v_0$ then labeled by $w_1 \bar u$, and since this path ends at $v_0$, then it may be followed by a path labeled by $w_2$. On the other hand it is easy to see that if the greed $\beta$-expansion is dense then each word has to appear in it infinitely often, so each word is a label appearing over infinitely many different paths in the graph representation. No such word can be a synchronizing word.
