A key property that all metric spaces have is the Hausdorff axiom. Namely given any $x,y \in X$ such that $x \neq y$ there are open sets $U$ about $x$, $V$ about $y$ such that $U \cap V = \emptyset$.
To see how this holds in metric space take any two distinct points $v$ and $w$. They are distinct so by definition of a metric $d(v,w) = \epsilon$ for some $\epsilon > 0$. By the Archimedean property of the reals there exists $n \in \Bbb{N}$ such that
$$ 0 < \frac{1}{n} < \epsilon$$
from which it follows that $B_{\frac{1}{n}}(v)$ does not intersect $B_{\epsilon - \frac{1}{n}} (w)$. Since $v,w$ were any two distinct points in your metric space this proves the claim.
So to show that your topological space above cannot be turned into a metric space it is sufficient to show that there are positive integers $x,y \in Z$ such that any two open neighbourhoods about $x$ and $y$ respectively must have non-trivial intersection. Consider $1$ and $2$ that are in $\Bbb{Z}$. Then it is easy to see that the only open neighbourhood about $1$ is $O_1 = \{1,2,3, \ldots \}$ while the only open neighbourhood about $2$ is $O_2 = \{2,3,\dots \}$ or even $O_1$ as well. However it is clear that
$$O_1 \cap O_1 \neq \emptyset, \hspace{5mm} O_1 \cap O_2 \neq \emptyset$$
from which it follows that the topology $\mathcal{J}$ that you put on $Z$ is not Hausdorff and hence is not metrisable.
You seem to have already made up your mind that this isn't worth your time, so it's hard to know what can be provided as an answer. But I would say that this is worth your time if for no other reason than you are clearly struggling to translate between two axiomatic systems that describe the same thing. That is a fundamental mathematical skill and anyone with a Ph.D. will tell you that they needed it, in some form or another, to get where they are.
I personally needed this definition, and at least a dozen other similar ones, for my thesis work. Let me try to explain why without getting into too much detail. The key point is this: Different sets of axioms generalize differently, even if the axioms themselves are equivalent. You are already starting to see this when you call axiom N5 "horrible"; indeed, in this way of viewing topology, filters are natural, but open sets are not. Obviously, in the most popular definition of a topological space, open sets are taken as fundamental.
Before 1990, rigid analytic geometry (which is very far from being a backwater) was largely studied by means of Grothendieck topologies, which are more or less a big categorical generalization of filters or neighborhood spaces. This is a bit strange, as there were actual topological spaces. But these spaces didn't behave in intuitive ways, for example they failed to be locally compact, an important technical property that went through just fine in the more abstract setting.
But Berkovich demonstrated that we can add some points (well, a lot of points) to get a really nice topological space, one that is locally compact and Hausdorff. Why was this possible, and why did nobody notice this before?
The key lies in something like axiom N5. You see, traditional non-archimedean spaces have, in a sense, too many open sets. The whole point of using Grothendieck topologies was to ensure that not every open set containing a point was a neighborhood of that point, by forcing it to contain some "much smaller" neibghborhood (in the literature this is sometimes called overconvergence).
Somehow, mathematicians knew that these open sets were not "open enough", so they had to be somehow excluded. Now we sort of know why: a (nice) classical non-archimedean space $X$ is a dense subspace of a locally compact Hausdorff space $\hat{X}$ (whose points may be difficult to get a handle on concretely). In particular, some open sets/coverings in $X$ are special, namely the preimages of open sets/coverings in $\hat{X}$. In fact, this means that the space $\hat{X}$ was always somehow implicit in a neighborhood system on $X$! (i.e. $\hat{X}$ consists of the points of some topos, blah blah blah)
For me, the moral of this story is that doing mathematics is not just about learning the most popular axioms and drawing conclusions from them. It's about understanding the interplay between different definitions. New research is often about forging new connections, and it's held back often precisely because some area is considered "backwater" and so on. Mathematics can go in or out of style, but it rarely becomes truly obsolete. Don't just chase style, take the time to understand the various perspectives (even the silly ones) and I can promise you it will pay off.
Best Answer
Yes, every neighborhood of $a$ must contain $a$; but it is not true that any particular neighborhood is a basis for the neighborhood system, because it may not be contained in every neighborhood. For example, consider the real line and $a=0$. The set $(-1,1)$ is a neighborhood of $a$, but is not by itself basis of the neighborhood system at $0$ because, for example, the neighborhood $(-1/3,1/2)$ does not contain the set $(-1,1)$.
A basis for the neighborhood system is a collection of neighborhoods with the property that any neighborhood contains (at least) one element of the system. They can be thought of as "small enough representatives" so that at least one of these representatives is contained in any given neighborhood.
First, note that $\mathcal{B}_a$ is a set of sets, not a set of elements of the space. Perhaps you mean $\mathcal{B}_a=\bigl\{\{a\}\bigr\}$, rather than $\mathcal{B}_a=\{a\}$. Second: if $\{a\}$ is a neighborhood of $a$ (it may not be: for example, it may not be open!) then it is true that one can take $\mathcal{B}_a$ to be just $\bigl\{\{a\}\bigr\}$. But for example, $\{\{0\}\}$ is not a basis of the neighborhood system for $0$ on the real line (with the usual topology), because $\{0\}$ is not even a neighborhood of $0$.
It is true that if there is a "smallest open set that contains $a$", then one can take that set alone to be a basis of the neighborhood system. But in many topological spaces, no such set exists. In the real numbers with the usual topology, there is no "smallest open set that contains $a$", so a basis of the neighborhood system at $a$ needs to have more than one element: given any open set that contains $a$, there is always a strictly smaller open set that contains $a$, so a basis for the neighborhood system in this topological space would necessarily have to contain infinitely many elements.
(An example of a basis of the neighborhood system at $a$ for the real numbers with their usual topology would be $$\mathcal{B}_a = \left\{ \left.\left(a -\frac{1}{n},a+\frac{1}{n}\right)\right|\, n\in\mathbb{N}\right\}$$ which you can verify: each element of $\mathcal{B}_a$ is a neighborhood of $a$, and every neighborhood of $a$ contains at least one element of $\mathcal{B}_a$; you can also replace the sets in $\mathcal{B}_a$ with the corresponding closed intervals, which are also neighborhoods and have the relevant properties.)