Consider linear functions over $\Bbb{R}$ of the form $f(x)=mx+b.$
This is visibly a vector space, and we may go further to obtain an associative algebra-like-object by giving this vector space a "multiplicative" binary operation defined by composition. This defines a left and right binary operation on this vector space.
That is $f_1 =m_1 x + b_1$ and $f_2 =m_2 x + b_2,$ then define
$$f_1\circ f_2 =m_1 m_2 x + m_1b_2+b_1.$$
Further, given $f_3=m_3 x + b_3,$ then
$$f_1\circ f_3 =m_1 m_3 x + m_1b_3+b_1,$$
and $$f_1\circ f_2+f_1\circ f_3 = m_1(m_2 +m_3)x+ m_1(b_2 +b_3)+2b_1.$$
Where
$$f_1\circ (f_2+ f_3)= f_1\circ[(m_2+m_3 )x+b_2+b_3] =m_1(m_2+m_3)x+m_1(b_2+b_3)+b_1 $$
so
$$[f_1\circ f_2+f_1\circ f_3]-[f_1\circ (f_2+ f_3)]=b_1.$$
Choosing any $b_1\ne0,$ yields an element $f_1$ which will not possess the distributive property.
So we have all the desirable properties of an algebra, without the left or right distributive property.
What you're describing is just a 1-dimensional vector space $V$ over $\mathbb{R}$. Since $V$ has dimension 1, any nonzero vector in $V$ is a basis, and so any other vector can be written as a scalar multiple of this vector (in light of the last condition you mentioned). The first two conditions are simply the requirements $V$ is closed under addition and scalar multiplication.
In fact, in physics, we often want to let the underlying field be complex; e.g. impedance in an RC circuit. So really the "precise name" you're looking for is a one dimensional vector space over $\mathbb{R}$ or $\mathbb{C}$.
Best Answer
My comment was a bit wrong, and too brief, so I'm going to expand this into a partial answer.
Suppose you have a set $R$ with operations $+$ and $\times$ such that $(R, +)$ is a group, and $\times$ left-endodistributes over $+$. Let $0$ be the identity under $+$.
Fix $a, b \in R$. Then $$a \times 0 = a \times (0 + 0) = 0 \times a + a \times 0 \implies 0 \times a = 0.$$ We also have $$0 = 0 \times (a + 0) = a \times 0 + 0 \times 0 = a \times 0,$$ and $$a \times b = a \times (b + 0) = b \times a + a \times 0 = b \times a.$$
Therefore, $\times$ is commutative, and therefore distributes over $+$.
The same can be done with left-exodistributivity.
For left-antidistributivity, consider first $$0 \times 0 = 0 \times (0 + 0) = 0 \times 0 + 0 \times 0 \implies 0 \times 0 = 0.$$ Next, note that $$0 \times a = 0 \times (a + 0) = a \times 0 + 0 \times 0 = a \times 0.$$ Then, $$(0 + a) \times (0 + a) = (0 + a) \times 0 + (0 + a) \times a = 0 \times 0 + 0 \times a + a \times 0 + a \times a,$$ which, when combined with the above identity, simplifies to $0 \times a + 0 \times a = 0$. But then, $$0 = 0 \times a + 0 \times a = a \times (0 + 0) = a \times 0 = 0 \times a.$$ Finally, this again gives us that $\times$ is commutative and distributes over $+$, as $$a \times (b + 0) = b \times a + 0 \times a = b \times a.$$
While this doesn't mean that left anti/endo/exo-distributivity properties are of no interest, it does mean that, in order to avoid "trivial" examples (i.e. ones where $\times$ distributes), we have to sacrifice a fair amount of structure of the additive magma, which means the result is not going to be as "ring-like" as you might have hoped.