What is the motivation for fixing boundaries on what is and is not a "distance". Sometimes one runs into multiple situations where the same type of theorems and proofs apply and then abstracts the pertinent properties which allow a theory. What to you actually want to do?
The article to which you link gives a particular meaning to dissimilarity which might be something like distance but not all the way to a distance. If it was exactly the same, why use a new word? I can imagine having distance $d(x,y)=0$ means close enough that it counts as the same. One can imagine situations where $d(x_i,x_{i+1})=0$ for $1 \le i \le n$ but $d(x_1,x_n) \gt 0.$ In that case, if in addition all distances are integral, we might have $d(x,z) \le d(x,y)+d(y,z)+1.$
Incidentally, sometimes a distance could fail to be symmetric, say in a directed graph.
I like thinking about the question but philosophically, what is the criterion for when we stop using distance? The universe can make a difference too. Suppose that we are on the unit sphere with ordinary points and we use the ordinary $\mathbb{R}^3$ Euclidean difference (We live on the surface but can run wires through the interior.) This seems like a pretty fair Space/distance pair. By my calculations $F(u,v)=u\sqrt{1-(\frac{v}{2})^2}+v\sqrt{1-(\frac{u}{2})^2}.$ For small $u,v$ this can beis just a bit less than $u+v$. It is associative but not scale invariant. Also, one could say $F(u,v)=\min(1,u\sqrt{1-(\frac{v}{2})^2}+v\sqrt{1-(\frac{u}{2})^2}).$ Then it is not associative.
As a perspective, what properties are required for a multiplication? Usually (but see below) a binary operation and it should help (some group of people who communicate with each other) to think of it as a multiplication. But when it doubt we say " a commutative and associative multiplication with cancelation" and spell out what can be assumed. Matrix multiplication is not always commutative. The cross product in $\mathbb{R}^3$ is not associative but it does distribute over vector addition. Multiplication of octonians is not associative but it is when restricted to the sub-algebra generated by any two elements and it extends the multiplication. There is the concept of a non-associative Algebra. so maybe that there is an addition to distribute over? Of course given a set with a multiplication operator we can define a formal addition which the multiplication should distribute over.
The distributive law says that the multiplication (in whatever context) is linear in each variable (with respect to the addtion). There is multilinear algebra with operators linear in each variable. The determinant is a multiplication of $n$ vectors in $\mathbb{R}^n.$ Given $n-1$ vectors $v_1,v_2,\cdots,v_{n-1}$ in $\mathbb{R}^n$ there is a unique $w \in \mathbb{R}^n$ so that for any $v_n$ the dot product $w \cdot v_n$ is the signed volume (determinant) determined by $v_1,\cdots,v_n.$ So here we have that $(v_1,\cdots,v_{n-1}) \mapsto w$ is a "generalized cross-product."