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).