Let two skew-symmetric matrices $A$, $B$ be given. We can apply the Cayley transformation $A\mapsto (I-A)(I+A)^{-1}$ to get orthogonal matrices which do not have an eigenvalue $=-1$.
Let $\tilde{A}:=(I-A)(I+A)^{-1}$ and $\tilde{B}:=(I-B)(I+B)^{-1}$ the Cayley transformations. Now, the matrix product $\tilde{A}\tilde{B}$ is again orthogonal and does not have an eigenvalue $-1$. So, there is a skew-symmetric matrix $C$ which fulfills $\tilde{A}\tilde{B}=(I-C)(I+C)^{-1}$. Is there a simple formula for $C$ in dependence of $A$ and $B$?
Yes! In fact, the inverse of Cayley transformation is itself a kind of Cayley transformation, and it turns out that
$$C = (I+\tilde{A}\tilde{B})^{-1}(I-\tilde{A}\tilde{B})$$
It's easy enough to prove this, so I'll write down the proof here. Let's start from the initial statement that
$$\tilde{A}\tilde{B}=(I-C)(I+C)^{-1}$$
From here, we can obtain $C$ using some algebraic manipulation.
$$\begin{align*} \tilde{A}\tilde{B}&=(I-C)(I+C)^{-1}\\ \tilde{A}\tilde{B}(I+C)&=I-C\\ \tilde{A}\tilde{B}+\tilde{A}\tilde{B}C&=I-C\\ \tilde{A}\tilde{B}C+C&=I-\tilde{A}\tilde{B}\\ (\tilde{A}\tilde{B}+I)C&=I-\tilde{A}\tilde{B}\\ C&=(I+\tilde{A}\tilde{B})^{-1}(I-\tilde{A}\tilde{B}) \end{align*}$$
Or, we can also express $C$ using the initial $A$ and $B$ matrices by using the first formula:
$$\begin{align*}C = &(I+\tilde{A}\tilde{B})^{-1}(I-\tilde{A}\tilde{B}) =\\ = &(I+(I-A)(I+A)^{-1}(I-B)(I+B)^{-1})^{-1}\ \cdot\\ \cdot\ &(I-(I-A)(I+A)^{-1}(I-B)(I+B)^{-1})\end{align*}$$