The speed of the car
It is correct to take "The speed of the car is $60$ km/h" and multiply through by $1$ hour to get "The ____ of the car in $1$ hour is $60$ km". However, it is not correct to call the result "speed". The thing that the car gets $60$ km of in $1$ hour is distance traveled. We can say "The distance the car travels in $1$ hour is $60$ km".
The price of the bottles
If you're buying bottles, you measure the total amount you pay in dollars; what you measure in dollars per bottle is "unit price" (not a terribly standard term). The goal of this "unit price" is to get a property of the thing you want to buy (water or milk or olive oil) that doesn't depend on how much you buy.
In the sentence "The price of $6$ bottles of olive oil is $6$ dollars" and divide through by $6$ bottle to get "The unit price of olive oil is $6$ dollars per $6$ bottles, or $1$ dollar/bottle". Think of this as an equation separated by "is"; you must do the same thing to both sides.
If you divide by $1$ bottle on both sides instead, you still have $6$ left on both sides, and so you have "$6$ times the unit price of olive oil is $6$ dollars/bottle". A true statement, but not a very useful one.
The mass of a mole of carbon
This is a super confusing one.
From one point of view, we're not doing any division. "Molar mass" and "Mass of $1$ mole" are simply synonyms. So we can say "The mass of $1$ mole of Carbon-12 is $12$ grams" or we can say "The molar mass of Carbon-12 is $12$ grams".
From another point of view, we'd like to be able to say things like
100 g of water is about 5.551 mol of water. (Source: Wikipedia)
It sure looks like we have a "mass of water" on one side, and a "number of water molecules" on the other side.
To enable doing this without making mistakes, we write molar mass in units of g/mol, so that we can use the molar mass of a substance to convert from moles to grams. If you want to turn "$1$ mole of Carbon-12" into "$12$ grams of Carbon-12", you multiply by the molar mass of Carbon-12 in grams/mole. That unit is chosen so that the units in
$$
1 \text{ mol} \cdot 12 \text{ } \frac{\text{g}}{\text{mol}} = 12 \text{ g}
$$
cancel out.
From a third point of view, a mole is simply a number of particles, so it's unitless(?) anyway.
Best Answer
The point of having units in computations is to restrict what one can do, such that does doesn't end up inadvertently writing down a computation whose result depends on one's choice of units in an unexpected way.
In mathematics this is not done a lot, but mathematics can certainly do it if we want to. What you do is to work with matrices (etc) over the field of rational functions of $n$ variables, and assign one of the variables to stand for each of your fundamental units. That is, we're working in the field of fractions of the polynomial ring $\mathbb R[\rm m, s, kg, A, \ldots]$.
The fraction field contains all of the (unitless) real numbers, as well as all of the unitful measuremens such as $2\,{\rm s}$ or $4000\,{\rm m}$ or $35\,\rm{\frac m{s^2}}$. It also contains a lot of nonsensical elements such as $\frac{42\,\rm{kg}}{3\,{\rm m}+8\,{\rm s}}$, but that doesn't really bother us, because we know that the field-of-fractions construction still keeps the entire system consistent.
We can then define that the computation we're speaking of is well-formed if we can prove that the output always can be written as a real vector (or matrix or whatever) times a scalar such as $1\,{\rm \frac ms}$, which defines the dimension we expect (on physical grounds) that the answer must have.
Once this proof (known to physicists and engineers as dimensional analysis) has been carried out, we can then use an evaluation homomorphism to map all of the unit-variables to $1$, so they disappear from the formulas and the actual calculation can be done on ordinary real numbers.
Whether this is exciting or not is subjective.