Is a linear vector space a vector space?
Here is a counterexample. Consider $\mathcal{X}=\{0,1\}$, with addition defined by $x+y=\max(x,y)$ and scalar multiplication defined by $ax=x$ for all $a\in\mathbb{R}$ and $x\in\mathcal{X}$. This satisfies all of Hale's axioms, but $1$ has no additive inverse.
Here's a slightly less trivial example. Consider the set $\mathcal{X}=\mathbb{R}\cup\{z\}$, with addition and scalar multiplication defined as usual for elements of $\mathbb{R}$, $x+z=z+x=x$ for all $x\in\mathcal{X}$, and $az=z$ for all $a\in\mathbb{R}$. This satisfies the given axioms, with $z$ as the zero element. However, no element other than $z$ has an additive inverse.
(In fact, any example without additive inverses contains a copy of the first counterexample. If $\mathcal{X}$ satisfies Hale's axioms and $x\in\mathcal{X}$ has no additive inverse, then $\{0,0\cdot x\}\subseteq \mathcal{X}$ will be closed under addition and scalar multiplication and isomorphic to the first example, sending $0\cdot x$ to $1$. We must have $0\cdot x\neq 0$ since $x+(-1)\cdot x=0\cdot x$ so $x$ would have an additive inverse if $0\cdot x=0$.)
Note, though, that additive inverses aren't the only axiom that is missing. Associativity of $+$ and $a(x+y)=ax+ay$ are missing too! Here's an example that has additive inverses but which fails associativity. Let $\mathcal{X}=\mathbb{R}\times\{0,1\}\setminus\{(0,1)\}$. We define addition by $(x,i)+(y,j)=(x+y,\max(i,j))$ and scalar multiplication by $a(x,i)=(ax,i)$, except that if either operation gives an output of $(0,1)$, we change it to $(0,0)$ instead (so for instance, $0(x,1)=(0,0)$ for any $x$). This satisfies Hale's axioms, and has additive inverses ($(-x,i)$ is the inverse of $(x,i)$). However, it fails associativity, since $$((x,0)+(-x,0))+(x,1)=(0,0)+(x,1)=(x,1)$$ whereas $$(x,0)+((-x,0)+(x,1))=(x,0)+(0,0)=(x,0)$$ for any $x\neq 0$.
The author almost certainly does not intend to give a different definition from the usual one, though--this is just an error in the book. It is definitely not standard to use the term "linear vector space" to refer to this weaker definition.
The definition is indeed missing something for a vector space, but I suspect that is not intentional. “Linear space” is a common synonym of “vector space”, probably because it is linear functions that respect the structure of a vector space.
To see that the conditions are not sufficient, consider $\mathcal X = \mathbb R\times \mathbb N$ with the addition $(a,m)+(b,n) = (a+b,\max\{m,n\})$ and the multiplication $a(b,n)=(ab,n)$.