The axioms of the list you cite are not independent, but $1x=x$ is not the problem.
Commutativity of addition follows from the other axioms; let $x,y\in V$ and set
$$
z=(1+1)(x+y)
$$
Then
$$
z=1(x+y)+1(x+y)=x+y+x+y
$$
(parentheses can be omitted because of associativity). On the other hand
$$
z=(1+1)x+(1+1)y=1x+1x+1y+1y=x+x+y+y
$$
Therefore, being $V,+$ a group, we can do
$$
(-x)+x+y+x+y+(-y)=(-x)+x+x+y+y+(-y)
$$
which gives
$$
y+x=x+y
$$
An correct objection would be that, removing the commutativity of addition axiom, only right zero and right opposites are assumed. However a general result about monoids applies.
Let $M$ be a (multiplicative) semigroup, with right identity $e$. If every element has a right $e$-inverse, then every element has a left $e$-inverse and the right identity $e$ is also a left identity.
The assumption is that $ae=a$, for all $a\in M$, and that, for all $a\in M$, there exists $b\in M$ such that $ab=e$.
Let $a\in M$ and $b\in M$ such that $ab=e$. Then $bab=be=b$, so, if $c\in M$ and $bc=e$, we have $babc=bc$, hence $ba=e$. Moreover, $ea=aba=a$. Thus $e$ is also a left identity and $b$ a left inverse of $a$.