Massive edit to simplify the question. Some comments below might be made obsolete - specifically, the comment that this follows directly from Dirichlet. That was true for the original wording.
I'm looking for a short proof, directly from Bézout's identity, of the following theorem:
Theorem 1: If $(a,b,c)=1$ then there exists an integer solution $x,y,z$ to $ax+bxy+cz=1$.
The case $(a,b)=1$ turns out to be equivalent to the theorem:
Theorem 2: The natural map: $$\mathbb Z_{nm}^\times\to\mathbb Z_{n}^\times$$ is onto.
That's because "onto" means if $(a,n)=1$ then for some $y$, $a+ny\in\mathbb Z_{mn}^\times$, meaning that $(a+ny,m)=1$ and thus $1=(a+ny)x+mz=ax+nxy+mz$ has a solution. The converse is equally obvious.
The general case in the first theorem follows if we know the case when $(a,b)=1$ since, for general $(a,b)$, we have $\left(\frac{a}{(a,b)},\frac{b}{(a,b)}\right)=1$, so from the special case, we get: $$\frac{a}{(a,b)}x_0 + \frac{b}{(a,b)} x_0y_0 + cz_0= 1$$ which implies:
$$ax_0 + bx_0y_0 + c((a,b)z_0)=(a,b)$$ Since $1=(a,b,c)=((a,b),c)$ we can find $(u,v)$ so that: $$(a,b)u + cv = 1$$
We then get:
$$a(x_0u) + b(x_0u)y_0 + c((a,b)z_0u + v) = 1$$
So $(x,y,z)=(x_0u,y_0,(a,b)z_0u+v)$ is a solution for our original equation. (Thanks Patrick Da Silva for that reduction.)
I can easily prove Theorem 2 using the structure of $\mathbb Z_n^\times$ in terms of prime factorizations. Indeed, Theorem 2 was the motivation for this question, initially - at first I thought it was "obvious," but the immediately realized it wasn't absolutely trivial.
It's certainly possible to translate the "abstract" proof of Theorem 2 into a direct proof of the special case of Theorem 1 using prime factorizations and Chinese remainder theorem.
But something about this theorem rang a bell for me. It looks like the sort of theorem that would have a short Bezout's identity proof.
Both unique factorization and Chinese remainder theorem are actually direct results of Bézout, and often theorems that we intuitively understand in terms of unique factorization and/or Chinese remainder theorem have a short, sharp proof using Bézout that eschews both the words "prime" and "remainder."
My instinct is that there ought to be a quick proof of the above with Bézout, without calling out to primes or remainders, but I haven't found it.
It's trivial if $(a,c)=1$, since $ax+cz=1$ lets us use $y=0$ to get a solution to $ax+bxy +cz=1$.
It's a little harder to see if $(b,c)=1$, but still not hard, since if $bu+cv=1$ then $$a\cdot 1 + 1\cdot (1-a) = a\cdot 1 + b(u(1-a)) + c(v(1-a))$$ giving a solution $(x,y,z)=(1,u(1-a),v(1-a))$.
That asymmetry (it's easy to solve if $(a,n)=1$ and harder to solve if $(b,n)=1$) suggests I might be wrong about there being such a proof, since Bézout is such a symmetric statement.
If there was a proof, it seems like you ought to start with:
$$au+bv=1\\ax+ny=(a,n)\\bw+nz = (b,n)$$
As an example of a theorem that is "obvious" with unique factorization, but has a simple proof with Bézout's identity, consider:
$(a,n)=(a,m)=1\implies (a,mn)=1$
That has a unique factorization proof, but it follows directly from Bézout by multiplying: $$1=(ax_1+ny_1)(ax_2+my_2) = a(ax_1x_2 + mx_1y_2+nx_2y_1) + mn(y_1y_2)$$
So, again, the goal is to have nothing about primes or Chinese Remainder Theorem in the proof, and to have it be "remarkably brief" - as much as possible, it shouldn't be hiding proofs of CRT or unique factorization.
I don't know that such a proof exists, but some instinct told me it did.