It's known that when $a$ and $b$ are comprime $\mathbb{Z}[i]/(a+bi) \cong \mathbb{Z}_{a^2+b^2}$, and $\mathbb{Z}[\omega]/(a+b\omega) \cong \mathbb{Z}_{a^2+b^2-ab}$, where $\omega$ is a primitive cube root of unity.
Is it true for rings $\mathbb{Z}[u]$ where we can define a norm $N$ that $\mathbb{Z}[u]/(a+bu) \cong \mathbb{Z}_{N(a+bu)}$?
If $u$ is an algebraic integer and $a,b\in \Bbb{Z},\gcd(a,b)=1$ then yes.
$a+bu$'s minimal polynomial is $f=\sum_{n=0}^d c_n x^n\in \Bbb{Z}[x]_{monic}$
$$(-1)^d N(a+bu) =f(0)=-\sum_{n=1}^d c_n (a+bu)^n\in (a+bu)$$
From the $\det$ definition of the norm we have $N(a+bu) = h(a,b)$ for some polynomial $h\in \Bbb{Z}[y,z]$ so that $$N(a+bu) \equiv N(a)\equiv a^n \bmod b$$ $$\implies \gcd(b,N(a+ub))=1,\quad \exists c, bc\equiv 1\bmod N(a+ub)$$
And hence $$\Bbb{Z}[u]/(a+bu)=\Bbb{Z}[u]/(a+bu,N(a+bu))=\Bbb{Z}[u]/(ca+u,N(a+bu))\cong \Bbb{Z}/(N(a+bu))$$ (isomorphism as rings)