For a simple example where (2) does not imply (1), let $R$ be any nonzero ring and let $S$ be the rng with the same underlying abelian group as $R$ but with the multiplication $ab=0$ for all $a,b\in R$. Consider the $S$-module $M=R\times R$, with $S$ acting by $a\cdot(b,c)=(0,ab)$. There is an epimorphism of $S$-modules $p:M\to S$ given by the projection onto the second coordinate, and $\ker p$ is isomorphic to $S$. The epimorphism $p$ cannot split, since otherwise we would have $M\cong S\oplus S$ and the action of $S$ on $M$ would be trivial. Thus $S$ does not satisfy (3) (or equivalently, (1)) as a module over itself.
For an example where (1) does not imply (2), as well as a broader understanding of projective modules over rngs, let us back up a bit and understand what a module over a rng really is. If $R$ is a rng, then there is a ring $U(R)$ obtained by freely adjoining a unit to $R$: the underlying abelian group of $U(R)$ is $\mathbb{Z}\oplus R$, and multiplication is given by $(n,a)(m,b)=(nm,nb+ma+ab)$. You should think of $(n,a)$ as being $n\cdot 1+a$ (since $(1,0)$ is the unit of $U(R)$).
It is easy to see that an $R$-module is in fact the same thing as a (unital) $U(R)$-module: given an $R$-module $M$, let $U(R)$ act on it by $(m,a)\cdot x=mx+ax$, and conversely, given a $U(R)$-module, just consider the action of elements of the form $(0,a)$. It is also easy to see that under this correspondence, a homomorphism of $R$-modules is the same thing as a homomorphism of $U(R)$-modules.
It follows that definitions (1) and (3) are the same thing for $R$-modules and $U(R)$-modules. Since $U(R)$ is unital, they also coincide with (2) for $U(R)$-modules. Thus we can say: an $R$-module $M$ is projective (i.e., satisfies (1)) iff as a $U(R)$-module, $M$ is a direct summand of a free $U(R)$-module. In particular, $U(R)$ (considered as an $R$-module) is always projective. But, for instance, if $R$ satisfies $ab=0$ for all $a,b\in R$, then $U(R)$ cannot be a direct summand of a free $R$-module (since $R$ acts nontrivially on $U(R)$). So for any rng $R$ with trivial multiplication, $U(R)$ is an $R$-module which satisfies (1) but not (2).