Let $a$, $b$ be positive, relative prime integers. Let $c$ be another positive integer such that $a \nmid c$, $b \nmid c$ and $ab − a − b < c < ab$. We want to show that $ax+by=c$ has unique solution for positive integers. My idea was that, given that $a$ and $b$ are coprimes, the solutions must be of the form $x=cx_0+bt$, $y=cy_0-at$, where $(x_0, y_0)$ is a solution for $ax+by=1$ and $t$ is any integer. If we set the condition that $x$ and $y$ must be nonnegative, we bound the possible values of $t$ in $[-cx_0/b, cy_0/a]$. From there on I've tried in different ways to see that this interval is narrower than $2$ using both $ax_0+by_0=1$ and $ab − a − b < c < ab$, but I don't really succeed.
$ax+by=c$ has unique solution for positive integers
diophantine equationsdivisibilityelementary-number-theorygcd-and-lcm
Related Solutions
If $(a,b)\mid c$, $\exists t\in \mathbb{Z}$ such that $t(a,b)=c$. As we know that there exists $x,y \in \mathbb{Z}$ such that $ax+by=(a,b)$, then choose the integers $x_0=tx$ and $y_0=ty$.
To prove the converse, if $x_0$ and $y_0$ be integers, then $(a,b)\mid x_0a$ and $(a,b)\mid y_0y$ implies $(a,b)\mid (x_0a+y_0b)$.
In your proof attempt, your point #$2$ doesn't give the set of all solutions (e.g., it has the parity of $z$ always being the same as that of $z_0$). Also, I don't offhand see how, but I believe it's fairly complicated, to try finishing using your approach.
Instead, note the question just asks to prove there exists an $n_0$, not necessarily to determine the smallest one. As you indicated, Bézout's identity states there exist integers $x_0$, $y_0$ and $z_0$ where
$$ax_0 + by_0 + cz_0 = 1 \tag{1}\label{eq1B}$$
Choose sufficiently large non-negative integers $i_1$, $i_2$ and $i_3$ so $x_1 = x_0 + i_1 \ge 0$, $y_1 = y_0 + ai_2 \ge 0$ and $z_1 = z_0 + ai_3 \ge 0$. Using these new values on the LHS increases the result by $ja$, where $j = i_1 + bi_2 + ci_3 \ge 0$, to now have
$$v_1 = ax_1 + by_1 + cz_1 = ja + 1 \tag{2}\label{eq2B}$$
Multiplying by $k$ for $1 \le k \le a$ gives
$$v_k = ax_k + by_k + cz_k = jka + k \tag{3}\label{eq3B}$$
where $x_k = kx_1$, $y_k = ky_1$ and $z_k = kz_1$.
Set $n_0 = v_a$. For any $n \ge n_0$, there's a $1 \le k \le a$ where $n \equiv k \equiv v_k \pmod{a}$. Thus, using \eqref{eq3B}, $x = x_k + \frac{n - v_k}{a}$, $y = y_k$ and $z = z_k$ are non-negative integers such that
$$ax + by + cz = n \tag{4}\label{eq4B}$$
Note if $a$ is not the smallest among $a$, $b$ and $c$, then using the smallest instead will usually give a smaller $n_0$. Also, the procedure above also shows how to prove there's an $n_0$ for any set of $4$ or more coprime coefficients.
FYI, regarding finding the smallest $n_0$, as stated in Coin problem,
If the number of coin denominations is three or more, no explicit formula is known.
Also, as indicated in Pete L. Clark's answer,
In the $k = 3$ case that you are asking about, an earlier paper of Harold Greenberg gives an algorithm which is simpler, and (if I am not mistaken) faster than that of the general case.
Best Answer
First you need to show that a solution exists for $c$ in the given range; that is done e.g. here.
Then because $c<ab$, $0\le x<b$ and $0\le y<a$. But the smallest vector (by norm) we can add to $(x,y)$ to generate another solution is, since $(a,b)=1$, $(-b,a)$. If we add this vector $x$ will become negative; if we subtract it $y$ will become negative, and both are disallowed. Thus the solution that we proved exists for $ax+by=c$ is unique.