Your matrix can be expressed in the form $M = P \otimes I$, where
$$
P = \pmatrix{a&b&\cdots\\b&a&b&\cdots\\
\vdots&\ddots&\ddots&\ddots}
$$
and $\otimes$ denotes the Kronecker product.
First, let's remark on when the matrix $M$ has a Cholesky decomposition. $M$ has a Cholesky decomposition if and only if it is positive semidefinite, and it turns out (as a consequence of the properties of the Kronecker product) that $M$ is positive semidefinite if and only if $P$ is positive semidefinite.
With that said, we'll find a Cholesky decomposition of $P$ (if $P$ is positive definite) and use this to get a Cholesky decomposition of $M$. As discussed here, it is relatively easy to show that the eigenvalues of $P$ will be equal to $a + (n-1)b$ (with multiplicity $1$) and $a-b$ (with multiplicity $n-1$). If both of these are positive, then one can compute a Choelsy decomposition for $P$.
Suppose that $L$ is a lower triangular matrix such that $LL^T = P$. It follows from the properties of the Kronecker product that $L \otimes I$ is lower triangular and that
$$
(L \otimes I)(L \otimes I)^T =
(LL^T) \otimes (II)^T = P \otimes I.
$$
That is to say, if $P = LL^T$ is the Choelsky decomposition of $P$, then $M = (L \otimes I)(L \otimes I)^T$ will be the Cholesky decomposition of $M$.
If you are simply looking for any matrix $R$ such that $M = RR^T$, then we can avoid numerically computing the Cholesky decomposition by using the eigenvalues/eigenvectors of $P$ to produce the (symmetric) positive semidefinite square root of $M$. To start, we will find an expression for the positive semidefinite square root of $P$. Let $J$ denote the $n\times n$ matrix
$$
J = \frac 1n\pmatrix{1 & 1&\cdots\\1&1&\cdots\\\vdots&\vdots&\ddots}.
$$
Notably, we can express $P$ as
$$
P = (a - b) I + nb J.
$$
Now, write
$$
Q = p I + q J.
$$
Using the fact that $J^2 = J$, we find that
$$
Q^2 = p^2 I + (2pq + q^2) J.
$$
Thus, if we select non-negative $p,q$ such that
$$
\begin{cases}
p^2 = a-b\\
2pq + q^2 = nb,
\end{cases}
$$
then we will find that $Q$ is positive semidefinite and that $Q^2 = P$. Solving these equations yields non-negative roots
$$
p = \sqrt{a-b}, \\
q^2 + 2pq - nb = 0 \implies q = \sqrt{a-b} + \sqrt{a + (n-1)b}.
$$
With that, we can show that the matrix $R = Q \otimes I$ is also positive semidefinite and satisfies $R^2 = M$.
