Q:

If 1 is the identity of the multiplicative (semi)group what is the term for 0?

Broadly given an operator $*$ the term identity is used for an element $e$ such that $x * e = x$ for all elements.  However is there a term for a value $ x * O = O$ for all values? This was brought to mind by this question What is the identity in the power set of $\Sigma^*$ as a monoid? that shows that the empty language has this property under concatenation. False has this property under the and operator.

A:

There is no such element in a nontrivial group.  Every element of a group has an inverse.
Let $G$ be a group and suppose (for purpose of contradiction) $G$ contains your proposed element $O \neq e$.  Then there is a $p = O^{-1} \in G$ such that $pO = e \neq O$.  But this contradicts the definition of $O$.  Therefore, there is no nontrivial group, $G$, containing an $O \neq e$ as described.
Another way to get at this, using the required existence of inverses, is that from
$$  x O = O  \text{,}  $$
we have
$$  x = xO O^{-1} = O O^{-1} = e  \text{.}  $$
So the assumed multiplication properties of $O$ are incompatible with its membership in a group unless the only element of the group is $e$ (in which case $e = O$ does satisfy the properties of both the multiplicative identity in a multiplicative group and the properties of the $O$ element you describe).  (This is why I wrote "$O \neq e$" in the second paragraph: to avoid the case that we were secretly only talking about the group with one element.)

A:

Excepting the special case of a group with only one element, groups cannot have $0$ as an element, and multiplication as its operation, as it is required that every element have an inverse, and what is the multiplicative inverse of $0$?
It sounds like you are interested in Rings.
Rings take a set with two binary operations, one operation is analogous to addition and the other is analogous to multiplication.  A ring has an additive identity ($0$) and an multiplicative identity ($1$) and requires that multiplication distribute over addition.  As consequence $a\cdot 0 = 0.$
(Yes, there is also the special case here, where the ring has one element.)
Rings have a generalization of $0$, which is called an "ideal."  An ideal is a subset of the ring such that for every element in the ideal multiplied by a member of the ring gives an element in the ideal.

