(I'm not sure this qualifies as an answer or an extended comment, because I don't have any deep knowledge in the matter, but I hope this can at least help clear some possible confusion.)
It seems to me that there are (at least) two different reasons to hypothesize the existence of some kind of “field with one element”, which may or may not be the same in both cases:
The first reason comes from the classification of the semisimple algebraic groups $G$: the Weyl group $W(G)$ of $G$ behaves in many ways similarly to the group of points of $G$ over a field that should have $1$-element (inter alia, $\# G(\mathbb{F}_q)$ is a polynomial in $q$, and $\# W(G)$ is the value at $1$ of that polynomial multiplied by $(q-1)$ to the appropriate power so it is nonzero; but also, the Weyl group has much structure, like parabolic subgroups and such, that is analogous to the $G(k)$ for $k$ a field).
So the idea there is that $G$ has more structure than being just an algebraic group over $\operatorname{Spec}\mathbb{Z}$: this means that (just like an algebraic variety $X$ over $\mathbb{Q}$ has more structure than the algebraic variety $X_{\mathbb{C}} := X \times_{\operatorname{Spec}\mathbb{Q}} \operatorname{Spec}\mathbb{C}$ that it defines) we can forget this extra structure by moving up to $\operatorname{Spec}\mathbb{Z}$, but by doing so we are discarding something. So $\operatorname{Spec}\mathbb{Z}$ should be thought of as being above $\operatorname{Spec}\mathbb{F}_1$, and by forgetting the combinatorial structure (root system, Weyl group, etc.) we move from $G$ over $\operatorname{Spec}\mathbb{F}_1$ to $G$ over $\operatorname{Spec}\mathbb{Z}$ (or over any ring in the usual sense).
It is not clear whether $W(G)$ should be taken literally as $G(\operatorname{Spec}\mathbb{F}_1)$ in this setup, but at least it should have some connection to it: again, the idea is that $G$ lives over $\operatorname{Spec}\mathbb{F}_1$ and you can pull it back to $\operatorname{Spec}\mathbb{Z}$ if you want, but you then lose the ability to compute $G(\operatorname{Spec}\mathbb{F}_1)$.
The second reason to hypothesize some kind of $\mathbb{F}_1$ comes from the analogy between function fields (of curves over finite fields) and number fields: so roughly $\operatorname{Spec}\mathbb{Z}$ should play a role analogous to $\operatorname{Spec}\mathbb{F}_q[t]$. The main obstacle in this analogy is that $\operatorname{Spec}\mathbb{F}_q[t]$ has a base field $\mathbb{F}_q$ whereas $\operatorname{Spec}\mathbb{Z}$ does not. So $\mathbb{F}_1$ is supposed to be that base field, and $\operatorname{Spec}\mathbb{Z} \to \operatorname{Spec}\mathbb{F}_1$ should be somewhat analogous to $\operatorname{Spec}\mathbb{F}_q[t] \to \operatorname{Spec}\mathbb{F}_q$.
That second picture of $\mathbb{F}_1$ is less clear than the first (to me at least; and again, it's not clear whether the two pictures really refer to the same object), in part because it's not clear what kind of $\operatorname{Spec}\mathbb{Z} \times_{\operatorname{Spec}\mathbb{F}_1} \operatorname{Spec}\mathbb{Z} \to \operatorname{Spec}\mathbb{Z}$ we should get by pulling back $\operatorname{Spec}\mathbb{Z} \to \operatorname{Spec}\mathbb{F}_1$ by itself, or, relatedly, what the $\mathbb{F}_1$-points of $\operatorname{Spec}\mathbb{Z}$ should be (if this makes any kind of sense); so I can't say whether there should be a morphism $\operatorname{Spec}\mathbb{F}_1 \to \operatorname{Spec}\mathbb{Z}$ (i.e., an $\mathbb{F}_1$-points of $\operatorname{Spec}\mathbb{Z}$) at all or not, but the structural map is in the direction $\operatorname{Spec}\mathbb{Z} \to \operatorname{Spec}\mathbb{F}_1$.
(Still, one possible idea in this second picture is that by forgetting the $\mathbb{F}_1$-structure on $\operatorname{Spec}\mathbb{Z}$ we lose its Archimedean place $\infty$ and associated Arakelov structure. But it's not clear that this really refers to $\operatorname{Spec}\mathbb{Z}$ or some kind of completion thereof, like $\mathbb{P}^1_{\mathbb{F}_q}$ is to $\operatorname{Spec}\mathbb{F}_q[t]$.)
