To show Playfair implies Parallel is the same as to show (not Parallel) implies (not Playfair). So assume (not Parallel), i.e. there is a line $t$ (extension of the segment you refer to) intersecting lines $l,l'$ and such that the sum of the interior angles on one of the sides of $t$ is less than $180^{\circ}$ (two right angles), yet $l,l'$ do not meet on that side of $t$. They cannot meet on the other side either, since on that other side the sum of interior angles exceeds $180^{\circ}$, and even in neutral geometry (i.e. with no parallel postulate) the sum of two angles in a triangle cannot exceed $180^{\circ}$. Therefore $l,l'$ are parallel. We may also construct another line $l''$ through the point $P$ on $l'$ where the transversal $t$ meets it, such that for this line the sum of its internal angles on the same side of $t$ is equal to $180^{\circ}$, and this line $l''$ is parallel to $l$ by neutral geometry only. So we now have two distinct parallels to $l$ through $P$ and have arrived at the negation of the Playfair axiom.
Both (i) and (ii) are strictly weaker than the parallel postulate. Probably the easiest way to see this is using coordinates. Given any Pythagorean ordered field $F$ and a convex subring $F_0\subseteq F$, the full subplane $F_0^2$ of the coordinate plane $F^2$ is a Hilbert plane which will always satisfy (i) and (ii), and will not satisfy the parallel postulate unless $F_0=F$. (If $F$ is non-Archimedean, you can always find such an $F_0$ which is not all of $F$, for instance by taking $F_0$ to be the set of all "finite" elements of $F$, i.e. elements bounded by integers.) Conversely, every Hilbert plane satisfying the parallel postulate is isomorphic to the coordinate plane over some Pythagorean ordered field, and so will satisfy (i) and (ii).
Here is a proof that $F_0^2$ satisfies (i) and (ii). For (i), suppose we have a non-vertical line with equation $y=ax+b$. If the slope $a$ is in $F_0$, then $ax\in F_0$ for all $x\in F_0$, and thus we must have $b\in F_0$ for the line to have any points in $F_0^2$ at all, in which case the line intersects every vertical line. On the other hand, if the slope $a$ is not in $F_0$ (or if our line is vertical), then $a$ must be infinitely large (since $F_0$ is a convex subring of $F$, it contains all elements which are bounded by an integer), so $a^{-1}$ is infinitesimal and is in $F$. Rewriting the equation as $x=a^{-1}(y-b)$, the argument from the first case now shows that our line intersects every horizontal line.
Thus, every line in $F_0^2$ either intersects every vertical line or intersects every horizontal line. Given three lines, we may thus assume without loss of generality that two of them intersect every vertical line. Now take any point on the third line, and the vertical line on that point will be a transversal to all three lines. (Unless the third line is itself vertical, in which case you can take its intersection with the first line and draw the line from it to any other point on the second line, choosing the latter point so that the line is not vertical.)
For (ii), suppose we have four (or indeed any finite collection of) points in $F_0^2$. Choose $a<b$ in $F_0$ such that both coordinates of all four points are strictly between $a$ and $b$. Then the right triangle with vertices $(a,a)$, $(a,2b-a)$, and $(2b-a,a)$ contains all four points. (Note that this argument actually requires only that $F_0$ is a subgroup of $F$, not a subring as was needed for (i).)
Best Answer
Your expression is indeed wrong, since $\forall l\forall P:P\notin l$ means that "no point is on a line". What you want is in fact: $$\forall l\forall P\big(P\notin l\to\exists !m\left(P\in m\land m||l\right)\big)$$ which translates as: "for any line $l$, if a point is not on this line, then there is a unique line $m$ to which this point belongs and which is also parallel to $l$".
Now, for the contradiction of this statement, recall that $\neg\forall x\equiv\exists x\neg$ and $\neg(A\to B)\equiv A\land\neg B$. However, it is slightly more complicated to negate $\exists !x$, and I refer you to the (excellent) accepted answer on this post. The idea is that, to contradict "there exists a unique $x$", we want to say "either there is no $x$ or there is more than one". Thus the negation of the statement is: $$\exists l\exists P\big(P\notin l\land\neg\exists !m(P\in m\land m||l)\big)$$ $$\exists l\exists P\Big(P\notin l\land\forall m\big(P\in m\land m||l\to\exists m'(P\in m'\land m'||l\land m'\neq m)\big)\Big)$$ which is precisely what you want, since it states "there is a line $l$ and point not on it, such that if a line $m$ goes through that point and is parallel to $l$, then there is another such line".
From the point of view of logic, this works if there is no line going through $P$ at all, or if none can be found to be parallel to $l$. But of course, geometrical considerations remove the need for these cases and yield your desired statement of "To a given line and a point not on it, there is more than one line through this point parallel to it".