I think this identity is true. It's quite similar to
this. I'll give a
combinatorial argument.
Let's try and count the number of permutations of $\{1, \dotsc, n - 1\}$ that do
not fix $1$.
One way to count is to note that the number of permutations fixing $1$ is just
the number of permutations of $\{2, \dotsc, n - 1\}$, so the answer is
$(n - 1)! - (n - 2)! = (n - 2)!(n - 2)$. (Or you can argue that there are $n - 2$ choices for where to send $1$, then $n - 2$ choices for where to send $2$, then $n - 3$ choices for where to send $3$...)
Another way to count is more complicated: for each $1 \le j \le n - 2$, choose
$j - 1$ elements of $\{2, \dotsc, n - 1\}$ to keep fixed (note a permutation
cannot fix all but one element), and then choose a
derangement of the remaining $n - j$ elements of $\{1, \dotsc, n - 1\}$. This
counts each of the permutations we're interested in exactly once, so it follows
that
\begin{equation*}
\sum_{j = 1}^{n - 2} c_{n - j} \binom{n - 2}{j - 1} = (n - 2)!(n - 2).
\end{equation*}
I can't find any references - since it's quite straightforward to prove, I don't
think there will necessarily be a really good reference, nor does it really require
one.