So, first of all we have $\mathbb{Z}_3=\{0,1,2\}$ and base vectors $\{e_0,e_1,e_2\}=\{[1,0,0],[0,1,0],[0,0,1]\}$ numerated by elements of group, which means we are looking for representation $\phi:\mathbb{Z}_3\rightarrow GL(3,\mathbb{C})$. Regular representation is given by linear transformation $\phi_s$ (for element $s\in\mathbb{Z}_3$) such that for given vector from base $e_r$ it returns $\phi_s(e_r)=e_{rs}$, so for element $0$, it does nothing (doesn't change base vectors) so it's given by matrix
$0\mapsto \begin{bmatrix}1,0,0\\0,1,0\\0,0,1 \end{bmatrix}$, $1$ shifts everything by one so first vector in base is now second etc. so
$1\mapsto \begin{bmatrix}0,1,0\\0,0,1\\1,0,0 \end{bmatrix}$, and $2$ shifts everything by two, so
$2\mapsto \begin{bmatrix}0,0,1\\1,0,0\\0,1,0 \end{bmatrix}$.
Is my reasoning with regular representation for given base correct?
I suspect that it will reduce into three $\mathbb{C}$ representations, because i know that if group is abelian, then its irreducible representations are of dimension 1, but i don't know how to get them.
You always have the trivial subrepresentation, given by the span of $e_1 + e_2 + e_3$.
Furthermore, to find them all, if an element is fixed by the generator of $\mathbb{Z}_3$, then it is fixed by $\mathbb{Z}_3$, so to find one dimensional $\mathbb{Z}_3$-invariant subspaces, this is equivalent to finding eigenvectors of the matrix of (what you have called) 1.
From this perspective, the trivial subrepresentation given above, corresponds to the eigenvector [1,1,1] with eigenvalue 1.
I hope this points you in the right direction!