A group-like structure with multiplicative zero instead of the identity My question is whether the structure in the title is known and has a name.
For clarity: the structure is a finite group like structure $G$ but instead of the identity we have a zero element $0\in G$ such that $0 g=0$ for all $g\in G$. 
The prototypical example of this structure is the subset of $n\times n$ matrices denoted with  $M_{ij}$  for $i,j=1,...,n$  (with matrix multiplication as the group product) and  defined by having $1$ in the $i,j$ position and $0$ everywhere else, and one additional matrix $M_0$ that is the all zeros matrix. 
 A: A set with a binary associative operation is a semigroup.
A semigroup that has a two-sided identity element is a monoid.
A monoid in which every element has a two-sided inverse is a group.
Given a semigroup $S$, an element $0\in S$ such that $0g=g0=0$ for all $g\in S$ is called a zero element.
You have "semigroups with zero" and "monoids with zero." Note that a zero element, if it exists, must be unique.
The set of $n \times n$ matrices with matrix multiplication are a (non-commutative) monoid with zero. Note that it does not, as you write, "have a zero instead of an identity." Rather, it has a zero in addition to having an identity (which is not a problem, since $0e = 0$). 
If your object has a zero but no identity, then it is a semigroup with a zero. An example of a semigroup with zero that is not a monoid could be the even integers under multiplication. 
The set of $n \times n$ matrices with a single $1$ and zeros elsewhere, plus the zero matrix, would be a semigroup with zero. 
A: Well, in each commutative ring $R$ (such as the ring of integers or the ring of square matrices), the zero element $0$ is absorbing, i.e., $r0 = 0 = 0r$.
Indeed, $0 = 0 + 0$ and so $r0 = r(0+0) = r0 + r0$. By adding the additive inverse $-r0$ to both sides (i.e., $r0 + (-r0) = 0$), $0 = r0$.
