Why are units called units? Why are units in abstract algebra called units? Is it just because they generalize the notion of $-1$ and $1$?, and the like? There's often a sense that units don't matter, when talking about things like irreducibility- is the name unit supposed to trivialize them somehow?
 A: "Unit" from "unity", from Latin unus meaning "one" (see uno, un, etc. in Romance languages). Thus, "1-like thing". In fact, even 1 itself is sometimes referred to as "unity", as in the term roots of unity.
A: In the Sciences a unit is a fixed amount of a quantity in question.
Every other amount is a multiple of the fixed amount.
In the Sciences the zero amount cannot be a unit because no other amount is a multiple of it. In rings, units are those elements with the property that all other elements are multiples of them. Just like in the Sciences.
A: 
To build upon what user467419 said: In science a unit is a fixed value for which every other amount of something is a multiple of that fixed value.
In rings, units are those elements with the property that all other elements are multiples of them. Specifically, if $u$ is a unit of a ring $R$, then there is a $v\in R$ such that $uv=vu=1$. Thus, for any $x\in R$, one can rewrite $x$ "in terms of the unit $u$" like so: $x=x(vu)=(xv)u$. Thus, $x$ is a multiple of $u$.

This can be understood well in seeing why a ring $F$ is a field (i.e. a commutative ring in which every nonzero element is a unit) if and only if its only ideals are $\{0\}$ and $F$.
Note every element in a field is a unit $u$. It follows that if we rewrite any element $x\in F$ in terms of that unit, we will get $x=ku$ for some $k\in F$. Now consider that any ideal is a kernel of a ring homomorphism $f:F\to R$. $f$'s kernel could be $\{0\}$, in which case $f$ is injective, and the ideal it corresponds to is $\{0\}$.
Otherwise, some element (and thus unit) $u$ maps to zero. Since we can write any $x$ as $x=ku$, that means $f(x)=f(k)f(u)=0.$ So the kernel is all of $F$.
Therefore the only ideals (homomorphism kernels) of a field are $\{0\}$ and $F$, because every element is a unit, so if one element goes to zero, they all must as well, since every element is a multiple of every other.

More generally, since every element in a ring can be written as a multiple of any unit, we can consider any $x$ in a ring $R$ to be $x=ku$, and the behavior of $x$ under any group homomorphism $g$ is determined by the behavior of $k$ and of $u$ under $g$, since $g(x)=g(k)g(u)$. This explains why $u$ is called a "unit" since we can measure behavior of element $x$ in terms of what coefficient $k$ is in front of $u$ when we express $x$ in terms of $u$.

Further (and I do not know if this has to do with how the term originated), a summary for the intuition of the term may be the following theorem:

Theorem (alternate definition of a unit): $u\in R$ is a unit iff for all $x\in R$, there exists a $v\in R$ such that $(xv)u=x$.

The fowards case of this theorem is proven above, while the backward case is provable letting $x=1$.
