I am trying to show that if a matrice $C$ is such that for all $\sigma \in \mathfrak{S}_n$, $$ \displaystyle \prod_{i=1}^n c_{i \sigma(i)} = 1 $$
then there exists diagonal matrices $A = \text{diag}(a_1, ..., a_n)$ and $B = \text{diag}(b_1, ..., b_n)$ such that $\det(A) = \det(B) = 1$, i.e. $$ \displaystyle \prod_{i=1}^n a_i = \prod_{j=1}^n b_j = 1 $$ and such that $c_{ij} = a_ib_j$ for all $i,j$.
This fact is implicit in the text I am reading (http://www.math.tamu.edu/~jml/yethesis.pdf for reference but it is hidden in the text) and is proven with Lie algebra, but my intuition is that this lone fact is much simpler to prove.
From the equations, we have that for all $i,j,k,l$, with $i \neq k$ and $j \neq l$, $$ c_{ij}c_{kl} = c_{il}c_{kj} $$ But I don't know what to do next. Can someone help me?
Whether this is true depends on the underlying field. It needs to contain a certain $n^{\rm{th}}$ root for $A$ and $B$ to exist. For example, the $2\times2$ real matrix $$ \begin{bmatrix} 1&-1\\ -1&1 \end{bmatrix} $$ satisfies the condition for $C$, but $A$ and $B$ would be (up to a choice of sign) $\operatorname{diag}(i,-i)$ and $\operatorname{diag}(-i,i)$.
Assuming we can take the root we want, here's how to construct $A$ and $B$. First, as you observed, the condition that the product of the elements $c_{i\sigma(i)}$ is independent of $\sigma$ means that, if $i \ne j$, we can replace $\sigma$ with a permutation that makes the opposite assignments to $i$ and $j$, leaving everything else the same. Only two factors in the product change, so $c_{i\sigma(i)}c_{j\sigma(j)} = c_{i\sigma(j)}c_{j\sigma(i)}$. More concretely, for any rectangle of elements in $C$, the two products of elements in opposite corners are equal.
Thus, for $i\ne1\ne j$, we have $c_{11}c_{ij} = c_{i1}c_{1j}$; equivalently, $c_{ij} = \dfrac{c_{i1}c_{1j}}{c_{11}}$. This is also trivially true if $i = 1$ or $j = 1$. This suggests how to proceed: Set $a_{i} = \dfrac{c_{i1}}{c_{11}}$ and $b_{j} = c_{1j}$. Then we have $c_{ij} = a_i b_j$, as desired. However, the products of the $a_i$ and the $b_j$ need not be $1$, although $\prod_i a_i \prod_j b_j = \prod_i c_{ii}$ is $1$. To fix this, simply divide each $a_i$ by an $n^{\rm{th}}$ root of their product and multiply each $b_j$ by that same root.