Your observations are correct and hold for arbitrary $m,n$. It suffices to note that
$$
A = \frac 1{mn}\pmatrix{
0&1&\cdots & 1\\
1&0&\ddots&\vdots\\
\vdots&\ddots&\ddots&1\\
1&\cdots&1&0} \otimes J_n + \frac 1m I_{mn}
$$
and use the properties of the Kronecker product.
In more detail: $C_{m \times m}$ is a rank 1 update of a scalar matrix, so we find that its eigenvalues are $-1$ with multiplicity $m-1$ and $m-1$ with multiplicity $1$. On the other hand, $J_n$ has eigenvalues $0$ with multiplicty $n-1$ and $n$ with multiplicity $1$.
It follows that $C \otimes J$ has eigenvalues $0$ with multiplicity $m(n-1)$, $-n$ with multiplicity $m-1$, and $n(m-1)$ with multiplicity $1$.
From there, it suffices to note that $\lambda$ is an eigenvalue of $A$ if and only if $c \lambda + d$ is an eigenvalue of $c A + dI$.
Let's address the easy part first: if the lower-left entry is a $0$ instead of $B$, then $C$ is block upper-triangular, which means that its determinant is given by the product of the determinants of the diagonal blocks. That is, we would find
$$
\det(C) = \det(A)^n = (a^2 - b^2)^n.
$$
In light of my below discussion of the original matrix: the eigenvalues of $A$ are of the form $a \pm b$. Correspondingly, $C$ has the same eigenvalues, but each with multiplicity $n$.
The matrix $C$ (as originally presented) can be written nicely in terms of the Kronecker product. In particular, let $J$ and $P$ denote the matrices (of sizes $2 \times 2$ and $n \times n$ respectively)
$$
J = \pmatrix{0&1\\1&0}, \quad P = \pmatrix{0&1\\&0&1\\&&\ddots&\ddots\\
&&&0&1\\
1&&&&0},
$$
and let $I_k$ denote the identity matrix of size $k$. Then
$$
C = aI + b(I_n \otimes J) + c(P \otimes I_2).
$$
The eigenvalues of $J$ are $\pm 1$, and the eigenvalues of $P$ are the $n$th roots of unity (i.e. $e^{2\pi ki/n}$ for $k = 0,\dots,n-1$).
The properties of the Kronecker product allow us to deduce that for matrices $X,Y$ with eigenvalues $\lambda_1,\dots,\lambda_m$ and $\mu_1,\dots,\mu_n$ respectively, it can be shown that the eigenvalues of $I_n\otimes X + Y\otimes I_m$ are given by $\mu_j + \lambda_k$ for all pairs $j,k$ with $1 \leq j \leq m$ and $1 \leq k \leq n$. Applying this to $X = bJ$ and $y = cP$ allows us to see that the eigenvalues of $C - aI$ are given by
$$
(-1)^jb + e^{2\pi k i/n} c, \quad 0 \leq j \leq 1, \quad 0 \leq k \leq n-1.
$$
Thus, the eigenvalues of $C$ have the form
$$
a + (-1)^jb + e^{2\pi k i/n} c, \quad 0 \leq j \leq 1, \quad 0 \leq k \leq n-1.
$$
The determinant of $C$ is the product of its eigenvalues.
If $a,b,c$ are real, it is notable that the eigenvalues of $C$ can be put into complex conjugate pairs. For $k = 0,1,\dots,\lfloor (n-1)/2\rfloor$, we find that
$$
(a + (-1)^jb + e^{2\pi k i/n} c)(a + (-1)^jb + e^{2\pi (n-k) i/n} c) = \\
|a + (-1)^jb + e^{2\pi k i/n} c|^2 = \\
[(a + (-1)^jb) + c\cos(2\pi k/n)]^2 + c\sin^2(2\pi k/n).
$$
With that, the determinant of $C$ can be written as the product of real numbers.
Best Answer
You can achieve this by multiplying the original matrix with $I \otimes C$ from the left.