Explanation for why $1\neq 0$ is explicitly mentioned in Chapter 1 of Spivak's Calculus for properties of numbers. During the first few pages of Spivak's Calculus (Third edition) in chapter 1 it mentions six properties about numbers.

(P1) If $a,b,c$ are any numbers, then $a+(b+c)=(a+b)+c$
(P2) If $a$ is any number then $a+0=0+a=a$
(P3) For every number $a$, there is a number $-a$ such that $a+(-a)=(-a)+a=0$
(P4) If $a$ and $b$ are any numbers, then $a+b=b+a$
(P5) If $a,b$ and $c$ are any numbers, then $a\cdot(b\cdot c)=(a\cdot b)\cdot c$
(P6) If $a$ is any number, then $a\cdot 1=1\cdot a=a$

Then it further states that $1\neq 0$. In the book it says that it was an important fact to list because there is no way that it could be proven on the basis of the $6$ properties listed above - these properties would all hold if there were only one number, namely $0$. 
Questions:
1) How does one rigorously prove that $1\neq0$ cannot be proven from the $6$ properties listed? 
2) It says that "these properties would all hold if there were only one number, namely $0$." Is a reason as to why this is explicitly mentioned is to avoid this trivial case where we only have the number $0$? Is there another deeper reason as to why this sentence was mentioned in relation to $1\neq 0$?
NB: Can someone please check if the tags are appropriate and edit if necessary? Thanks.

To prove that $1 \neq 0$ can't be proven from those properties, one can just construct an example where $1 = 0$ and those properties hold, since this means that you have structures that satisfy the axioms where some have $1 = 0$ and others have $1 \neq 0$, so this property is independent of the axioms.
Specifically, the zero ring (what you get if you have a single number) satisfies all of them. If you want, you can easily check manually since the only choice for any variable is $0$:
P1) $0 + (0 + 0) = 0 = (0 + 0) + 0$
P2) $0 + 0 = 0 = 0 + 0$
etc.
P6 is the one we need to look at specifically, it can be written as "There exists a number $x$ such that $a \cdot x = x \cdot a = a$, and we call this $x$ the $1$ of the ring". In the zero ring, choose $x = 0$, so
$0 \cdot x = 0 \cdot 0 = 0 = x \cdot 0$
Thus $0$ satisfies the rule for the "$1$" in the ring, and so $0 = 1$.
