Is there a simple example of a morphism of schemes $X\rightarrow S$ that is
Consider the map suggested by Boa. For any integral domain $R$, the fraction field with its usual $R$-module structure is flat. So the second bullet point is satisfied (alternatively, the second bullet point follows from the first because formally smooth morphisms between locally Noetherian schemes are flat).
For any point $p\in\mathrm{Spec}\:\mathbb{Z}$, the base change of the morphism $\mathrm{Spec}\: \mathbb{Q}\rightarrow \mathrm{Spec}\:\mathbb{Z}$ is either the unique morphism from the empty scheme so vacuously smooth (this is the case iff $p$ is a closed point) or it is an isomorphism (this is the case iff $p$ is the generic point). So the third bullet point is satisfied.
The set-theoretic image of the morphism $\mathrm{Spec}\:\mathbb{Q}\rightarrow \mathrm{Spec}\:\mathbb{Z}$ is not open so it can not be locally of finite presentation. So the fourth bullet point is satisfied.
Let $P$ be the set of positive prime numbers, think of $\mathbb{Q}$ as a $\mathbb{Z}$-algebra on generators indexed by $P$ and with the relations $px_p=1$ for all $p \in P$. Let $B$ be a commutative unital ring with a square-zero ideal $I$. Assume we have a morphism $\psi:\mathbb{Q}\rightarrow B/I$. Since the map $\pi:B\rightarrow B/I$ is surjective, for any prime $p$ we can find an element $a$ in the inverse image of $\psi(x_p)$ such that $pa=1+i$ for some $i\in I$. Set $a'=a(1-i)$. $a'$ belongs to the inverse image of $\psi(x_p)$ and satisfies $pa'=pa(1-i)=(1+i)(1-i)=1$.
Once the map is defined on the generators, it extends to monomials by multiplicativity and to the whole algebra by additivity, so we have a homomorphism $\phi:\mathbb{Q}\rightarrow B$ that satisfies $\pi\circ\phi=\psi$.
Therefore, $\phi$ solves the lifting problem, i.e. the map $\mathbb{Z}\rightarrow \mathbb{Q}$ is formally smooth. So the first bullet point is satisfied.