If a set of vector fails to have an additive identity, can it still have an additive inverse? Let $V = \mathbb{R}^2$
Let addition in $V$ be defined as:
$(u, v) + (x, y) = (u + x, 0)$
Let scalar multiplication be defined as:
$a(u, v) = (au, av)$
Clearly, there is no additive identity for $V$. Can $V$ still have an additive inverse, where we define the zero vector arbitrarily?
For example, could I define the zero vector in $V$ to be $(1, 0)$. Then the additive inverse would be $(-u + 1, a)$ where $a$ can be any real number:
$(u, v) + (-u + 1, a) = (1, 0)$
 A: Not in any usual sense, no. Note first of all that your structure is not a vector space, because it is in particular not a group... But you probably realize this already.
You can in theory define a "zero vector" however you want, but normally, a zero element is required to be an additive neutral element, i.e. $x+0=x$ and $0+x=x$, or at least one of those equalities should hold (right-neutral, left-neutral). There is no choice of "vector" in your structure that would satisfy this, so you're out of luck.

That was the boring answer... So what nice things can we say? Well, $V$ has commutative and associative addition, as well as scalar multiplication that almost work. What we lack is a zero vector (and thereby also additive inverses), as well as one of the distributivity laws for scalar multiplication:
$$
(a+b)(x,y) \ne a(x,y) + b(x,y) \quad\textrm{for}\quad y\ne0.
$$
The only sensible definition of a "pseudo-zero vector" in $V$ is probably $\mathbf 0 := (0,0)$. It at least has the nice properties that
$$
\mathbf 0+\mathbf 0=\mathbf 0 \quad\textrm{and}\quad
a\mathbf 0 = \mathbf 0,
$$
as well as the special cases $(x,0)+\mathbf 0=(x,0)$. With this we have non-unique additive inverses:
$$
(x,y) + (u,v) = \mathbf 0 \iff (u,v) \in \{(-x, s) \mid s\in\Bbb R\}.
$$
But what could we do with this? We would have for example
$$
\mathbf u + \mathbf v = \mathbf w
\implies \mathbf u + \mathbf 0 = \mathbf w - \mathbf v,
$$
for any choice of $-\mathbf v$. Thus, we can sort of do algebra, as long as we remember that $\mathbf u + \mathbf 0 \ne \mathbf u$ in general. We also have the nice fact that $(-1)\mathbf v$ is a possible choice for $-\mathbf v$.

Finally, I should point out that we have the obvious injection
$$
\varphi:\Bbb R \hookrightarrow V: x\mapsto (x,0).
$$
This acts like a homomorphism (i.e. a linear map), meaning
$$
\varphi(x+y)=\varphi(x)+\varphi(y)\quad \textrm{and} \quad \varphi(ax)=a\varphi(x).
$$
So $V$ has the vector space $\mathbb R$ embedded in it. Note also that $\varphi(0)=(0,0)$, which is kind of nice.

EDIT. Inspired Mozibur Ullah's answer, the additive structure of $V$ is in fact a commutative semigroup. In the language of semigroups, $(0,0)$ is the unique idempotent element of $V$, which is why it was the only natural choice of $\mathbf 0$. An idempotent element is always the identity element of a unique maximal subgroup; in this case, the maximal subgroup is $\{(x,0)\} \cong \mathbb R$. In particular, $\{(x,0)\}$ is the unique maximal subgroup of $V$.
A: The problem here is that a vector space has an additive structure which is defined to be an abelian group and this requires an identity, usually written as zero.
If we had an abelian structure that allowed for subtraction but without an identity we can see if we can build a vector space like structure over it.
Well, there is such a structure and its known as an inverse semigroup which like all semigroups is associative but has no identity. It is inverse because it does have unique inverses for every element in the sense:

for all $x$, there is a unique $y$ such that $x = xyx$ and $y=yxy$.

We can see why these axioms were chosen by supposing our inverse semigroup to actually be a group and then we see that $y$ is must be $x^{-1}$. Since being an inverse semigroup does not preclude it from having an identity, it may have one. We insist upon it.
Now, we want this to be abelian, so the axioms become:

for all $x$, there is a unique $y$ such that $x = x+y+x$ and $y=y+x+y$

However, to build an analogue of a vector space you would need to give an action of the reals (or of some field) on this semigroup. Again we hit a problem. This time about what to do when we multiply an element of the semigroup by the real zero. We have no zero in the the semigroup to set it to.
One option is to drop the real zero. This means we need to explore what a field looks like when it has no zero. We can copy what we did above and stipulate that its additive structure is an abelian inverse semigroup with no zero and we leave the multiplication as a monoid and ask that it distributes over addition. There is no standard name for this structure but for the purposes of this post, we could call it a semiring.
Now, in a field we have inverses for all non-zero elements. A semi-ring has no zero,  so we should ask that all elements are invertible. Again, there is no standard name for this structure, so we call it for the purposes of this post, a semi-field.
Finally, we define a semi-vector space to be an action of a semi-field on an additive inverse semigroup. This has no zero by construction.

Another possibility are unpointed true convex cones.
A cone is a subset of an ambient vector space which is closed under strictly positive scalings. This may include a zero, in which case we call it pointed. If not, we call it unpointed.
It is a true cone when for any $v$ in the cone, $-v$ doesn't belong to the cone. It is convex when it is closed under convex combinations. This is equivalent in the presence of scalings to be closed under strictly positive linear combinations. These are called conical combinations.
Geometrically speaking, this is a cone in the vector spaces with the tip on the origin but not including it.
A: Yes. Generalizing what you have said, if the zero vector is $(z,0)$, then for any vector $(u,v)$, an inverse of that vector is of the form $(z-u,a)$ where $a$ is any real number. Therefore every vector has infinitely many inverses.
