Proof

Let $x_0,y_0$ be a solution of the linear Diophantine equation (LDE) $ax+by=c.$
Set $x:=x_0+h\frac b{\gcd(a,b)}$ and $y:=y_0-h\frac a{\gcd(a,b)}$ for any $h\in\mathbb Z.$
Then $ax+by=ax_0+b_y=c.$

Assume, $x,y$ is any solution of the above LDE. We will show that it has always the required form.
By assumption, $ax+by=c=ax_0+by_0.\label{eq:E18579}\tag{1}$
We can assume $b\neq 0,$ otherwise we can exchange $a$ and $b.$ Furthermore, by the existence part, $\pm h$ are allowed in the formulae $x_0+h\frac b{\gcd(a,b)}$ and $y_0-h\frac a{\gcd(a,b)}.$
From $(\ref{eq:E18579}),$ it follows $$\begin{array}{rcl}(ax)(|b|)&\equiv&c(|b|)\\(ax_0)(|b|)&\equiv&c(|b|)\end{array}\label{eq:E18579a}\tag{2}.$$
With the existence and number of solutions of an LDE with one variable, it follows that $(\ref{eq:E18579a})$ has exactly one solution of the form $$x\equiv x_0\mod \frac{|b|}{\gcd(a,b)},$$ where $\gcd(a,b)$ is the greatest common divisor of $a$ and $b.$
Therefore, we can write $x=x_0+h\frac bd$ for some "integer:https://www.bookofproofs.org/branches/definition-of-integers/ $h\in\mathbb Z.$
It follows $$\begin{array}{rcl}by&=&c-ax\\ &=&c-a\left(x_0+h\frac b{\gcd(a,b)}\right)\\ &=&(c-ax_0)-b\frac{ha}{{\gcd(a,b)}}\\ &=&by_0-b\frac{ha}{{\gcd(a,b)}}\\ &=&b(y_0-h\frac{a}{{\gcd(a,b)}}. \end{array}$$
It follows that any solution has the claimed form.
∎

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Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927