Throughout, I'll work with your "short" axioms. I'll begin by addressing your questions about independence of the axioms.
First, the axiom $||1||\leq 1$ does not follow from the other axioms. To see this take $R=\mathbb{Z}$, let $c>1$ be a real number, and let your norm be given by $||x||=|x|\cdot c$ (where by $|x|$, I mean the usual absolute value of $x\in \mathbb{Z}$).
Second, the axiom $|-x|=|x|$ does not follow from the others. To see this again take $R=\mathbb{Z}$, let $c_1\neq c_2>1$ be real numbers with $c_2^2\geq c_1$, and let your norm by given by $$||x||=\begin{cases}|x|\cdot c_1 & \text{ if }x\geq 0\\ |x|\cdot c_2 & \text{ if }x<0\end{cases}.$$ A quick check should show this satisfies all of the other axioms, if I didn't make any mistake.
Third, I believe that Jérôme Poineau's comment is incorrect, in that by dividing by $||1||$ you can lose the norm property. To see this, take $R=\mathbb{Z}[y,y^{-1}]$, let $c>1$ be a real number, and let your norm be given by $$||\sum_{\rm finite}a_m y^m|| = \sum_{m<0}|a_m|+\sum_{m\geq 0}|a_m|\cdot c.$$ Again, if I didn't make any mistakes, this should satisfy all the norm axioms except that $||1||=c>1$. However, if you try to create a new norm $||f||'=||f||/||1||$, this loses the multiplicative property since $$||y\cdot y^{-1}||'=||1||'=1> 1\cdot c^{-1}=||y||'\cdot ||y^{-1}||'.$$
As for your question about which definition is preferred, I'm not an expert and will leave that to others.