I will try to provide an answer not along the pathologies that arise when removing axiom 3, but rather what a useful generalization should be.
Let me first remind you of the Veblen-Young Theorem:
If Desargues theorem holds in an abstract projective plane, it is of the form $P(V)$ for some vectorspace $V$ over a skew-field $k$.
If Pappus theorem holds as well, the field can be found to be commutative.
This is a form of concreteness result. In fact if we allow ourselves to talk about abstract projective spaces of general dimension, all abstract projective spaces of dimension greater than $2$ are of the form $P(V)$ for a vectorspace $V$ over a skewfield. This obviously requires axiom 3, since it holds for all projective spaces of the form $P(V)$!
As an example, the above theorem would obviously not hold if we allow as our projective plane a three element set with lines defined as two-element subsets. Both Pappus and Desargue are true in this 'projective plane'.
There is however a way of making sense of the above example: It "should" be a projective space over the "field with one element". Now note that such a thing in the strict sense of the wording doesn't exist. But in a lot of different areas in mathematics there's evidence that a generalization of the term field should exist and that there should be something people call 'absolute geometry', i.e. algebraic geometry over the field with one element.
If you want to see a definition of projective space that doesn't enforce axiom 3, but is still sensible, I recommend Cohn's introductory paper Projective Geometry over $\mathbb F_1$. Definition 1 in this paper may give you a more satisfactory feeling.
Suppose by contradiction there exists a line $a$ passing through a single point $A$. By axiom 3 there exist two other points $B$ and $C$ and by axiom 1 they belong to a line $r$ parallel to $a$.
By axiom 1 there exists a line $s$ passing through $A$ and $B$ and by axiom 2 there exists a line $t$ passing through $C$ and parallel to $s$.
Line $t$ does not pass through $A$, so $t$ is parallel to $a$. But then we have two different lines $r$ and $t$ passing through $C$ and parallel to $a$, which contradicts axiom 2.
Best Answer
As far as I can see, a line with two points satisfies this system of axioms.
None of these axioms postulate the existence of noncollinear points, but that is normally a feature of axioms for the projective plane and projective $3$-space.
Perhaps the author has given these axioms in addition to some others that occurred earlier?