Is there a name for this quaternion-like algebra? $\bullet\ \textbf{The Algebra in question}$
So if you take the ring of integers appended with the cube root of unity, $R=\mathbb{Z}[\zeta_3]$, it seems like you can construct this weird non-commutative algebra, $H=R[u,v]$, with the properties
$$u^2+u+1=v^2+v+1=0\quad\text{and}\quad vu = \zeta_3uv.$$
It would seem at first that $1,u,v,$ and $uv$ form the minimal basis of $H$ over $R$. But with some calculation,
$$v^2u=-vu-u=-\zeta_3uv-u\quad\text{and}$$
$$v^2u=\zeta_3^2uv^2=-\zeta_3uv^2-uv^2=\zeta_3uv+\zeta_3u+uv+u,$$
one can show that 
$$(-2\zeta_3-1)uv=(\zeta_3+2)v$$
and therefore that $uv=\zeta_3v\in\text{span}_R(1,u,v)$. Or something similar. So $[H:R]=3$?
$\bullet\ \text{Motivation}$
I was trying to construct something like the quaternions but with a norm of $a^3+b^3+c^3$ instead of $a^2+b^2+c^2+d^2$. I'd started out with $u^3=v^3=1$ but switched to aforementioned equations to reduce the basis. And indeed the construction seems successful. If we take $\phi_i:u,v\rightsquigarrow \zeta_3^iu,\zeta_3^iv$ then we get for $\alpha=a+bu+cv$ that
$$\alpha\cdot\phi_1(\alpha)\cdot\phi_2(\alpha)=a^3+b^3+c^3.$$
I'd suspect that this isn't exactly a multaplicative norm but it's interesting enough to investigate.
$\bullet\ \text{Question}$
So what even do you call a structure like this? The search "noncommutative ring extension" doesn't yield anything helpful -- at least, nothing helpful that I can make sense of. I've been reading about Banach and Clifford algebras with no luck. I've also read Conway's quaternion/octonion book but don't have it available at the moment.
 A: Consider the (associative) non-commutative polynomial ring $H=R\langle u,v\rangle/I$, where the ideal $I$ is generated by $u^2+u+1,v^2+v+1,vu-\zeta_3 uv$. (I find it is better to use $\langle$ $\rangle$ when the variables do not commute.) The ring $H$ is clearly spanned over $R$ by $1,u,v,uv$.
Next you compute $v^2u$ in two ways to deduce $(1-\zeta_3^2)u=(\zeta_3^2-\zeta_3)uv$, and hence that $3u=3\zeta_3^2uv$. You can also similarly compute $vu^2$ in two ways to deduce $(1-\zeta_3^2)v=(\zeta_3^2-\zeta_3)uv$, and hence that $3v=3\zeta_3^2uv$. Thus $3u=3v$, so from $vu=\zeta_3uv$ you get $3u^2=3\zeta_3u^2$, and hence $9u^2=0$. Then, from $3u=3\zeta_3^2uv$ you get $9u=0$.
So, once you invert 3 in $R$, then $u=v=0$ and $H=R$.
A: You want $H$ to be an $R$-algebra generated by $u$ and $v$ over $R$, as an algebra. Assuming $H$ is associative, it is a quotient of the algebra $\bar H:=R\langle u,v\rangle/I$, where $R\langle u,v\rangle$ is the ring of polynomials in two non-commuting variables $u$ and $v$ with coefficients in $R$, and $I$ is the two-sided ideal generated by your relations;
$$I:=(u^2+u+1,v^2+v+1,vu-\zeta uv).$$
For future convencience set $\varepsilon:=1-\zeta\in R$.
Observation 1: In $\bar{H}$ we have $\varepsilon u=\varepsilon v=\zeta^2\varepsilon uv$.
Proof.  In the quotient $\bar H=R\langle u,v\rangle/I$ we have
\begin{eqnarray*}
u+uv&=&u(1+v)=-uv^2=-\zeta v^2u&=&\zeta (1+v)u=\zeta u+\zeta vu&=&\zeta u+\zeta^2uv,\\
v+uv&=&(1+u)v=-u^2v=-\zeta vu^2&=&\zeta v(1+u)=\zeta v+\zeta vu&=&\zeta v+\zeta^2uv,
\end{eqnarray*}
and comparing these two shows that
$$
(1-\zeta)u=-(1-\zeta^2)uv=(1-\zeta)v.
$$
Note that $-(1-\zeta^2)=\zeta^2(1-\zeta)$, so this already shows that $\varepsilon u=\zeta^2\varepsilon uv=\varepsilon v$.$\hspace{10pt}\square$
Observation 2: In $\bar{H}$ we have $3=0$ and $\varepsilon u=\varepsilon v=\varepsilon uv=\varepsilon$.
From the identities $\varepsilon u=\zeta^2\varepsilon uv=\varepsilon v$ it follows that
\begin{eqnarray*}
\varepsilon u
&=&\zeta^2\varepsilon uv
=\zeta\varepsilon vu
&=&\zeta\varepsilon u^2
&=&-\zeta\varepsilon(u+1),\\
\varepsilon v
&=&\zeta^2\varepsilon uv
&=&\zeta^2\varepsilon v^2
&=&-\zeta^2\varepsilon(v+1),
\end{eqnarray*}
which shows that
$$
\zeta\varepsilon=-(1+\zeta)\varepsilon u=\zeta^2\varepsilon u
\qquad\text{ and }\qquad
\zeta^2\varepsilon=-(1+\zeta^2)\varepsilon v=\zeta\varepsilon v,
$$
and hence that $\varepsilon u=\zeta^2\varepsilon$ and $\varepsilon v=\zeta\varepsilon$. Then
$$\zeta\varepsilon^2
=\zeta(1-\zeta)\varepsilon
=\zeta\varepsilon-\zeta^2\varepsilon
=\varepsilon v-\varepsilon u=0,$$
because we saw in observation 1 that $\varepsilon u=\varepsilon v$. As $\zeta\varepsilon^2=-3\zeta^2$ it follows that also $3=0$ in $\bar{H}$. Moreover, this shows that $\zeta\varepsilon=\varepsilon$ and so $\varepsilon uv=\zeta^2\varepsilon uv=\varepsilon u=\varepsilon v=\zeta\varepsilon=\varepsilon$.$\hspace{10pt}\square$

Now we can reframe the question; we have seen that $\{1,\zeta,u,v,uv\}$ is a $\Bbb{Z}$-basis for $\bar{H}$, but multiplication in $\bar{H}$ is significantly simpler on the $\Bbb{Z}$-basis $\{1,1-\zeta,1-u,1-v,(1-u)(1-v)\}$.
Set $x:=1-u$ and $y:=1-v$. Observation 2 shows that $3,\varepsilon(u-1),\varepsilon(v-1),\varepsilon(uv-1)\in I$, and so $\varepsilon x,\varepsilon y,\varepsilon xy\in I$. Because $3\in I$ we see that $\bar{H}$ is an $R/3R$-algebra, where
$$R/3R\cong\Bbb{Z}[X]/(X^2+X+1,3)\cong\Bbb{F}_3[X]/(X^2+X+1)=\Bbb{F}_3[X]/(1-X)^2.$$
This shows that $R/3R\cong\Bbb{F}_3[\varepsilon]$ where $\varepsilon^2=0$. Of course also $u^2+u+1,v^2+v+1\in I$, where
$$u^2+u+1=(1-u)^2=x^2
\qquad\text{ and }\qquad
v^2+v+1=(1-v)^2=y^2,$$
and so $x^2,y^2\in I$. A routine check shows that $yx-xy+\varepsilon\in I$, so $\bar{H}/(\varepsilon)$ is commutative; in fact
$$\bar{H}/(\varepsilon)\cong\Bbb{F}_3[x,y]/(x^2,y^2).$$
This is a commutative ring of $3^4=81$ elements, and $H$ all its quotients are isomorphic to
$$\Bbb{F}_3[x,y]/(x^2,y^2),\qquad \Bbb{F}_3[x,y]/(x,y)^2,\qquad\Bbb{F}_3[x]/(x^2),\qquad\Bbb{F}_3,\qquad\text{ or }\qquad 0.$$
That is to say, if $\varepsilon=0$ in $H$, or equivalently if $\zeta=1$ in $H$, then $H$ is isomorphic to one of these five commutative rings. Otherwise, if $\varepsilon\neq0$ in $H$ then $H$ is isomorphic to a quotient of $\bar{H}$ by a two-sided ideal not containing $\varepsilon$. It turns out that every non-zero two-sided ideal of $\bar{H}$ contains $\varepsilon$:
Observation 3: If $J\subset\bar{H}$ is a nonzero two-sided ideal then $\varepsilon\in J$.
Proof. Let $J\subset\bar{H}$ be a two-sided ideal not containing $\varepsilon$ and let $h\in J$. Let $a,b,c,d,e\in\Bbb{F}_3$ such that
$$h=a+b\varepsilon+cx+dy+exy.$$
Then clearly $\varepsilon h=a\varepsilon\in J$, so $a=0$. Because $xy-yx=\varepsilon$ in $\bar{H}$ we see that $J$ contains
$$xh-hx=(d+e)\varepsilon
\qquad\text{ and }\qquad
hy-yh=(c+e)\varepsilon,$$
so $c+e=0$ and $d+e=0$. Then $h=b\varepsilon+c(x+y-xy)$ and so
$$h(1+y)(x-y)=(b\varepsilon+c(x+y))(x-y)=c(yx-xy)=-c\varepsilon,$$
hence also $c=0$. Then $h=b\varepsilon\in J$ and so $h=0$, which shows that $J=0$.$\hspace{10pt}\square$
In conclusion, there is only one such non-commutative algebra, up to isomorphism, and it is the finite ring
$$\Bbb{F}_3[\varepsilon]\langle x,y\rangle/(x^2,y^2,\varepsilon x,\varepsilon y,yx-xy+\varepsilon,),$$
of order $3^5=243$ in which $\varepsilon x=\varepsilon y=\varepsilon xy=0$.
