For non-representable morphisms of Artin (i.e. algebraic) stacks, different properties are defined in different ways, depending on their particular nature.
E.g. for properties which are smooth local on source and target, see here in the Stacks Project.
The property of being etale is not smooth local on source and target, so this approach doesn't work in that case. (If we restrict to DM stacks, we can use the etale analogue of this definition, and so in that case we get the correct definition of etale; see David Carchedi's answer.)
This doesn't mean that we can't define the notion of etale morphisms, though; it just means that this particular framework doesn't apply.
One approach is via infinitesimal lifting properties: using these one can define what it means for a morphism of stacks (representable or not) to be formally smooth, formally etale, or formally unramified. By imposing locally fin. pres. in the first two cases, or locally fin. type in the third case, one then gets notions of smooth, etale, and unramified.
Alternatively, one can define a morphism to be unramified iff it is loc. finite type and has etale diagonal. (Since diagonal morphisms are always representable, we know what etale means for the diagonal.)
We can then define a a morphism to be etale if it is flat, loc. fin. pres., and unramified.
Note that one can also define a smooth morphism using the framework of smooth-local-on-source-and-target discussed above, because being smooth is smooth local.
See Appendix B of this paper by David Rydh for a discussion of these various definitions and their equivalences. (Note though that the affineness condition on the schemes involved in the infinitesimal lifting properties seems to have been accidentally omitted.)
Note also that in the non-representable context, etale is stronger than smooth and quasi-finite, or smooth and relative dimension zero. (Since smooth implies flat, etale is equivalent to smooth and unramified.)
E.g. if G is a positive dimensional smooth alg. group over Spec $k$, then $$BG := [ \mathrm{Spec} \, k / G] \to \mathrm{Spec} \, k$$ is smooth and quasi-finite, but not etale. It is of negative relative dimension (equal to $- \dim G$.
The morphism
$$ [\mathbb A^1/ \mathbb G_m] \to \mathrm{Spec }\, k$$ is smooth and of relative dimension zero, but it is not etale.
(An etale morphism is unramified, thus has etale diagonal, thus has unramified diagonal, and thus is a DM morphism. In particular, if the target is a scheme, then the source is a DM stack, which $BG$ (for positive dimensional $G$) and $[\mathbb A^1/\mathbb G_m]$ are not.)
Added later: here is a correct characterization of etale morphisms of Artin stacks as certain kinds of smooth morphisms (akin to the fact that etale morphisms of schemes are smooth morphisms that are locally quasi-finite).
Recall that a morphism of Artin stacks is called Deligne--Mumford if it has unramified diagonal, or equivalently (but non-obviously) if it's base-change over any scheme yields a Deligne--Mumford stack (see here in the Stacks Project, especially footnote 1). (So this is a weakening of representability ---
which is equivalent to the diagonal being a monomorphism --- and is automatic for morphisms between DM stacks.)
Then a morphism $X \to Y$ is etale iff it is smooth, Deligne--Mumford, and locally quasi-finite.
(For the proof: note that an unramified morphism has (by definition) an etale, and so unramified, diagonal, hence is Deligne--Mumford. Using this,
one can also check that an unramified morphism is locally quasi-finite, because this reduces to checking that a Deligne--Mumford stack which is unramified over the Spec of a field is locally quasi-finite --- which is
easy. Since etale morphisms are in particular smooth, we get the only if direction.
For the if direction, by pulling back over a chart of $Y$ we may assume that $Y$ is a scheme, and hence that $X$ is a DM stack. Now we have to
check that a smooth and loc. quasi-fin. morphism from a DM stack to a scheme is etale, which is again easy.)
Yet another formulation of the same notion, adopting the view-point expressed by user t3suji in a comment on another answer:
A morphism $X \to Y$ of Artin stacks is etale if it is DM, and if for any morphism $V \to Y$ whose source is a scheme, the base-changed morphism
$$X \times_Y V \to V$$
(which is now a morphism from a DM stack to a scheme) is etale (in the sense of morphisms of DM stacks, where we already know what etale means: choose an etale chart $U \to X\times_Y V$, and require that the composite $U \to V$ be etale).
(The equivalence with the various preceding definitions is easily checked.)
Best Answer
To answer question 2, the best example I know is $\mathscr{M}_1$, the stack of (proper smooth geom. connected) curves of genus 1. Indeed, Raynaud has contructed an elliptic curve $E\to S$ over a scheme $S$ and an $E$-torsor $X\to S$ which is (an algebraic space but) not a scheme.
This implies two things. First, in order to define $\mathscr{M}_1$ we are forced to take "curve" to mean "algebraic space in curves": if we insist that curves must be schemes, the resulting $\mathscr{M}_1$ will not be a stack for the flat topology, because the above $X$ is locally a (projective) scheme for the flat (even étale) topology on $S$.
Concerning the diagonal, put $I:={\underline{\mathrm {Isom}}}_{\mathscr{M}_1}(E,X)$. There is a monomorphism $X\to I$ (in fact, a closed immersion) identifying $X$ with the subsheaf of $E$-torsor isomorphisms. So, $I$ is not a scheme because $X$ isn't. Viewed as a morphism $S\to \mathscr{M}_1$, $I$ is the pullback of the diagonal under the morphism $S\to \mathscr{M}_1\times\mathscr{M}_1$ given by $(E,X)$. Hence the diagonal is not representable in the scheme sense.