What properties make $[0,1]$ a good candidate for defining fundamental groups?

The answer to 1 is yes. For the purpose of this answer, a bipointed space is a topological space $J$ equipped with distinct closed points $e_0$ and $e_1$. As you say, for any bipointed space $J = (J, e_0, e_1)$, we can form a new bipointed space $J \vee J$ by taking the disjoint union of two copies of $J$, identifying the first $e_1$ with the second $e_0$, and giving the resulting space the obvious pair of basepoints.

Theorem: In the category of bipointed spaces $J$ equipped with a map $J \to J \vee J$, the terminal object is the bipointed space $([0, 1], 0, 1)$ equipped with the map "multiplication by 2" from $[0, 1]$ to $[0, 1] \vee [0, 1] \cong [0, 2]$.

Or informally: $[0, 1]$ has the structure needed in order to be able to define and compose paths, and is universal as such.

The theorem is proved here, and is a variant of a result of Peter Freyd's (which characterized the interval set-theoretically and order-theoretically, but not topologically). The idea that $[0, 1]$ is universal with the structure needed for homotopy theory is expanded on in these talk slides.

Rather than answering the question, I want to claim that it's the wrong question.

There is a way to define fundamental group(oid)s and other homotopical notions that makes no reference to the unit interval or any other bipointed space. This goes by names like "shape theory" or "Cech homotopy" or "toposic homotopy". The basic idea is that a "path" consists of a chain of overlapping open sets---a purely topological notion. There's a nice paper by John Kennison called "What is the fundamental group?" which approaches this idea in an elementary way (no toposes required).

This sort of "homotopy" can "see" things which homotopy defined in terms of the unit interval cannot. For instance, the Warsaw circle has trivial $\pi_1$ in the traditional sense, but Cech homotopy can see that it contains a loop. Same goes for the "long circle" (close up the long line into a circle).

One can then simply observe that if your space admits lots of maps into it from the unit interval (e.g. is locally path-simply-connected or whatever), then the fundamental group(oid) as defined in terms of open sets can equivalently be defined using maps out of the unit interval. So from this point of view, there is nothing fundamental about the unit interval; it's just that a lot of the spaces we care about do indeed contain lots of "paths", so that for them we can give a simpler definition of homotopy in terms of paths.

I only have a partial answer for 1. and a hopefully non-confusing answer to 2.

To start with, let us work with the fundamental groupoid, which is more, ahem, fundamental and better suited to generalisation. In particular, we can consider the set $\pi^J(X,a,b)$ of homotopy classes (rel endpoints) of maps $(J,0,1) \to (X,a,b)$, which is more natural in the setting you outline. This is, assuming it isn't empty, a torsor for the groups $\pi^J(X,a,a)$ and $\pi^J(X,b,b)$, so you're not really losing too much. But the more important structure is the whole groupoid.

The unit interval is at least weakly initial in the category of path-connected bipointed spaces and homotopy classes of maps (and we always have a torsor as above). If you don't assume path-connected, then the two-point set (with any of its topologies) can be allowed, but is completely useless in measuring homotopy. This is an important fact using $[0,1]$, and this can't be derived from formal homotopy theory. One could define $J$-connectedness for other bipointed spaces $J$, but the utility of such a definition is debatable unless you put in extra conditions, like making it a cylinder object.

The 'reason' we get a fundamental groupoid is that $[0,1]$ is an $A_\infty$ topological cogroupoid, namely a groupoid object in $Top^{op}$, up to homotopy, and then coherence of that up to homotopy, and so on, all the way up. Woah, I hear you say, that's a bit extreme. But it is true, and we can just focus on the first few layers.

First, we have a cocomposition $[0,1] \to [0,1]\sqcup_{1,0}[0,1]$ and a coidentity $[0,1] \to \ast$. Then instead of coassociativity, which would be the equality of the to obvious maps $$ [0,1] \to [0,1]\sqcup_{1,0}[0,1]\sqcup_{1,0}[0,1], $$ we have a homotopy between these two maps. We also have a map $$ [0,1] \sqcup_{1,0}[0] \to [0,1] $$ expressing the identity on the right, and a similar one on the left. Again, these aren't equal to the identity maps of $[0,1]$, but are homotopic to them. And again, we have coinverses up to homotopy. The choices of all these homotopies aren't important (although you can look up representatives in any book on algebraic topology), because the spaces of such homotopies are contractible.

When we want to involve another space and actually get $\Pi_1(X)$, what we do is hom this topological $A_\infty$-cogroupoid into the space $X$, and get an $A_\infty$-groupoid, and then we truncate it to a groupoid, by quotienting out by these homotopies that we have chosen (but remember the choices are unimportant). It is important that $[0,1]$ is path-connected, because this makes the $A_\infty$-cogroupoid contractible in certain technical ways which are important for generalisations to higher categories (most of the ideas in this answer come from Todd Trimble's work). For instance, in my thesis I defined a certain sort of fundamental bigroupoid which could be applied to topological stacks, and I relied heavily on the $A_\infty$-cogroupoid structure, because it was the only way I could prove I even had a fundamental bigroupoid (I confess I did have much more complicated interval objects than here).