How is rearranging $56\times 100\div 8$ into $56\div 8\times 100$ allowed by the commutative property?

Notice that you have always $\div 8$, no matter the order of the other terms. You don't divide by a different number. It might help to think $\div c$ as a multiplication with $d=1/c$. Then everything would look easier: $$a\times b\div c=a\times b\times d=a\times d\times b=a\div c\times b$$

Hope this makes sense.

$$ a\times b ÷ c $$

$$=a\times b\times\dfrac{1}{c}$$

$$=(a\times\dfrac{1}{c})\times b$$

$$=\dfrac{a}{c}\times b$$

$$=a÷c\times b$$

Others have answered the direct question, in that multiplication is commutative and that applies also to multiplication by a reciprocal (the equivalent of division). However there is an issue here with associativity and division which I think is worth mentioning. This has to do with the order in which operations are carried out.

So $a\div b \times c$ is being interpreted from left to right as $(a\div b)\times c=\cfrac {ac}b$, but done from right to left $a\div (b\times c)=\cfrac a{bc}$ and the two results are not the same.

Likewise with $a\div b \div c$ we have $(a\div b)\div c=\cfrac a{bc} \neq a\div (b\div c)=\cfrac {ac}b$.

So the conventional assumption that multiplication and division are operations of equal status and are carried out from left to right does make a difference in these cases and changes the result. This may be what is feeding your intuition that there is a potential problem with order.