What is $(\operatorname{SL}_2(\mathbb{R})\times\operatorname{SL}_2(\mathbb{R})\times\operatorname{SL}_2(\mathbb{R}))/\{\pm(I_2,I_2,I_2)\}$? I know that $\operatorname{SL}_2(\mathbb{R})/\{\pm\operatorname{I}_2\}=\operatorname{PSL}_2(\mathbb{R})$ and that $(\operatorname{SL}_2(\mathbb{R})\times\operatorname{SL}_2(\mathbb{R}))/\{\pm(\operatorname{I}_2,\operatorname{I}_2)\}\simeq\operatorname{SO}_o(2,2)$ where $\operatorname{SO}_o(2,2)$ is the connected component containing the identity of $\operatorname{SO}(2,2)$.
So my question is:
Do we have such an isomorphism for $(\operatorname{SL}_2(\mathbb{R})\times\operatorname{SL}_2(\mathbb{R})\times\operatorname{SL}_2(\mathbb{R}))/\{\pm(I_2,I_2,I_2)\}$?
 A: 
Do we have such an isomorphism for $(\operatorname{SL}_2(\mathbb{R})\times\operatorname{SL}_2(\mathbb{R})\times\operatorname{SL}_2(\mathbb{R}))/\{\pm(I_2,I_2,I_2)\}$?

Well, let's think about what the phrase "such an isomorphism" even means. Your two examples:
$$ \mathrm{PSL}_2\mathbb{R}=\frac{\mathrm{SL}_2\mathbb{R}}{\{\pm I_2\}}, \qquad \frac{\mathrm{SL_2\mathbb{R}\times\mathrm{SL}_2\mathbb{R}}}{\{\pm(I_2,I_2)\}}\cong \mathrm{SO}_0(2,1). $$
The first is effectively the definition of $\mathrm{PSL}_2\mathbb{R}$. In general, if $G$ is a classical linear group (defined as some set of matrices), then the projective version $PG$ is defined to be $G/Z(G)$, where $Z(G)$ is a subgroup of central diagonal scalar matrices. This applies for orthogonal, unitary, special, general linear groups and more. Thus, the equation for $\mathrm{PSL}_2\mathbb{R}$ is not an example of an exceptional isomorphism.
We can fix this though. There are two exceptional isomorphisms:
$$ \mathrm{SL}_2\mathbb{R}=\mathrm{Spin}(2,1), \qquad \mathrm{SL}_2\mathbb{R}\times\mathrm{SL}_2\mathbb{R}=\mathrm{Spin}(2,2). \tag{$\ast$}$$
The first has an interpretation in terms of the celestial sphere, projective geometry, and stereographic projection. Both have an interpretation in terms of "regular representations."
These are a particular kind of exceptional isomorphism called an accidental isomorphism, which is an isomorphism between a classical linear group and a spin group. A spin group is a certain double cover of the (identity component) of a special orthogonal group. In 3D, this can be represented with the so-called belt / string / plate tricks. The formal definition of spin groups involves generalizing quaternions with Clifford algebras. Typically instead of giving an explicit isomorphism to the spin group, a double covering homomorphism to a special orthogonal group is given.
The dimension of $\mathrm{SO}(p,q)$ is $\binom{n}{2}$, where $n=p+q$. There is no solution to this for $n=9$, so there cannot be an accidental isomorphism involving $(\mathrm{SL}_2\mathbb{R})^3$. (In fact, I think $n=4$ is the only time $\mathrm{SO}$ isn't indecomposable mod $\pi_0$, so to speak.) I am not familiar with any other kind of exceptional isomorphism that would suggest $(\mathrm{SL}_2\mathbb{R})^3$ is special.
In the case of quaternions $\mathbb{H}$, the unit quaternions ("versors") form a $3$-sphere and may be denoted $\mathrm{Sp}(1)$, or sometimes $\mathrm{U}(1,\mathbb{H})$ in analogy with unit reals ($S^0\cong\mathrm{O}(1)$) or unit complex numbers ("phasors," $S^1\cong\mathrm{U}(1)$, and we can have $\mathrm{Sp}(1)$ act on $\mathbb{R}^3$ (the imaginary quaternions in $\mathbb{H}$) by conjugation, or have it act on all of $\mathbb{H}$ by both left and right multiplication, yielding accidental isomorphisms
$$ \mathrm{Sp}(1)\cong\mathrm{Spin}(3), \qquad \mathrm{Sp}(1)\times\mathrm{Sp}(1)\cong\mathrm{Spin}(4). \tag{$\circ$}$$
By interpreting $\mathbb{C}\cong\mathbb{R}^2$ we can say $\mathrm{U}(1)\cong\mathrm{SO}(2)$, and by interpreting $\mathbb{H}\cong\mathbb{C}^2$ we can say $\mathrm{Sp}(1)\cong\mathrm{SU}(2)$. Thus, often sources will call the group of unit quaternions $\mathrm{SU}(2)$, even though this isn't strictly true.
The split quaternions are isomorphic (as an algebra) to $\mathbb{R}(2)$ (in this context, this is the algebra of $2\times2$ real matrices), and the unit split quaternions correspond to $\mathrm{SL}_2\mathbb{R}$. Doing the same things for split quaternions that I said we do with quaternions to get $(\circ)$, we reproduce the accidental isomorphisms $(\ast)$.
All of the spin groups $\mathrm{Spin}(p,q)$ with $p+q\le6$ have an accidental isomorphism with classical groups, but it's kind of like a patchwork quilt. There are different classical groups, over different fields (reals, complex numbers, quaternions), with different explanations that cover different signatures. (I've heard there's also an accidental isomorphism for signature $(6,2)$, which I haven't investigated.) These can be generalized with octonions, projective geometry, and exceptional Lie groups.
