Are there any typically studied structures (say $\mathfrak D$) where the additive identity, $0_\mathfrak D$ has a multiplicative inverse? Are there any typically studied structures (say $\mathfrak D$) where the additive identity, $0_\mathfrak D$ has a multiplicative inverse?
In most usual settings, i.e, complex and real numbers as fields, ring of matrices, etc, this isn't true, what about non-field/ring structures?
E: I'll try to be more clear, I'm looking for structures that may be less known, not fields/rings.
 A: Actually there is one way to save this, even in a "ring" setting. If $R$ is a ring, then extend $R$ by appending the symbol $\infty$ which is to be the multiplicative inverse of $0$. You then have to define how $\infty$ plays with every other element. The obvious thing to do is to require that $a.\infty = \infty$ for any $a\neq 0$ and $a+\infty = \infty$ for any $a$. Note that this forces $\infty$ to not have an additive inverse.
This still does not fix the issue raised by Reveillark. Since you still want a group structure going on underneath (since you're talking about $0$), you still have that $0.x = 0$ for $x\neq\infty$. The real issue is in the second line of his answer:
$$ 1 = \infty.0 = \infty.(0.x) = (\infty.0).x = 1.x = x.$$
There is only one place where we have any possible leeway and that is in the equality $\infty.(0.x) = (\infty.0).x$. The associativity of multiplication with respect to $\infty$ ruins the ring structure (and forces you into conclude exactly as Rev did). The way to save this is not to allow associativity of multiplication when it comes to $0$. We will still have associativity for any other $x,y\neq 0$ since $\infty.(x.y) = \infty = (\infty.x).y$ by definition.
In summary, it can be done, you just have to break associativity a little bit by not allowing $\infty$ to have an additive inverse and not allowing associativity of multiplication with $0$ and $\infty$ to make it work. I'm not sure what you'd call such a space, maybe a rigged ring or something.

Note that if you allowed $\infty$ to have an additive inverse, say $-\infty$, then you'd break associativity of addition since you'd get
$$a+(\infty-\infty) = a = (a+\infty)-\infty = 0.$$
This structure can actually be found in measure theoretic contexts since there they allow values of $\infty$ and $-\infty$ but not simultaneously (as you lose all structure that way).
A: I don't think this is a definite answer, but it shows that with a few assumptions regarding the structure of $\mathfrak{D}$, then the set becomes quite trivial.
Consider a "structure" $(\mathfrak{D},+,·)$. Let $0$ be the additive identity and $1$ be the multiplicative identity. Assuming right distributivity, we have that, for any $x\in D$:
$$0x=(0+0)x=0x+0x$$
Assuming that $(\mathfrak{D},+)$ is a group, it follows that $0x=0$. 
Let $a$ be the multiplicative inverse of $0$. Then, for any $x\in \mathfrak{D}$, if $·$ is associative:
$$1=a0=a(0x)=(a0)x=1x=x$$ 
Hence $D=\{0\}$, which isn't a very interesting object.
