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What would be a way to show that all straight lines can be expressed as the points satisfying $ax+by+c=0$ and vice versa?

Is this possible simply from the axioms of Hilbert or Euclid?

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    All you need is Thales' theorem (I mean: Intercept theorem). – 2017-02-21
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    @Aretino How so? Could you link literature performing this proof? – 2017-02-21

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By one of the axioms of Euclidean geometry, two distinct points define a unique line. If you plug distinct points into your equation, you don't get a unique solution, but instead a one-dimensional space of solutions: an equation and all its multiples. You can consider this an equivalence class, as usually done for homogeneous coordinates.

For the converse, you'd have to verify that every such equation contains at least two distinct points. You could pick those for $x=0$ and for $x=1$, but you'd have to provide some alternate handling for $b=0$. In that case you could swap the roles of $x$ and $y$, and hence of $a$ and $b$ as well. But what if $a=b=0$? Then you don't have a line. For $c=0$, every point will satisfy the equation, while for $c\neq0$ no (finite) point satisfies the equation. So you must take care to exclude the case $a=b=0$.

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    I think that verifying that every such equation contains at least two distinct points is not enough: you should also prove that all the other points are aligned with those two. Same remark for the first part: plugging two points into the equation is not enough, in my opinion. – 2017-02-22