The shortest euclidean distance between three points namely 1,2,3 follows that
where $dist(x,y)$ is the vector between points x and y.
Now we from daily experience know that space in itself without time is euclidean.
Now this linearity relation should be carried over to space-time. Why ?
Because if there are three simultaneous events for an observer , then their space-time distance must equal the euclidean distance and thus will follow the linearity condition.
So we expect that
The space-time interval between any events a,b and c must also follow the relation
$dist^* (a,c)=dist^* (a,b)+dist^* (b,c) $
where $dist^*$ is the space-time interval distance vector.
which will be followed only if the unit of space interval is length and not length$^2$.
This linearity relation also makes the math easier and lets us do things in special relativity similar to pre-relativity days like defining velocity,kinetic energy,momentum in a similar way newton did and they follow the same kind of vector/scalar addition respectively the way they did in pre-relativity days.
Now if you still did everything the same way like defining momentum to be
$p=$ $m$ x (new metric) $/$ proper time
energy to be having units $p^2/2m$.
Where new metric denotes the metric to be $\Delta s'$ in the question.
You won't be having the relations like energy conservation,momentum conservation to hold true in the same mathematical form they used to do earlier in pre-relativity mechanics.
So either treat minkowski metric to be of the dimensions $length$ or completely change the way you defined momentum,energy and everything before relativity so that your theory remains consistent with the universe.
The latter seems a very daunting task to do than the former.
To summarise : .
Our equations retain their old pre-relativity mathematical form is the core reason why we take space-time interval to be of the units of length.
Also our distance still is a vector quantity.