I don't know how satisfying this answer will be, but I'll give it a shot anyway. The punchline, I think, is that although these "diagonal properties" have just as much aesthetic appeal as their "anti-diagonal" counterparts, the diagonal properties happen to give us information that is more useful for a matrix as it is used mathematically. That is, diagonal symmetry is a more natural thing to look for in the context of linear algebra.
First of all, note that all of these properties are properties of square (that is, $n \times n$) matrices, which are implicitly linear maps from $\Bbb F^n$ to $\Bbb F^n$ (that is, they produce vectors of $n$ entries from vectors of $n$ entries).
The properties that we really care about in linear algebra are the ones that tell us something about how matrices interact with vectors (and ultimately, with other matrices).
Diagonal Matrices
Diagonal matrices are important because they describe a particularly nice class of linear transformations. In particular:
$$
\pmatrix{d_1\\&d_2\\&&\ddots \\ &&& d_n} \pmatrix{x_1\\ x_2\\ \vdots \\ x_n} =
\pmatrix{d_1 x_1\\ d_2 x_2\\ \vdots \\ d_n x_n}
$$
I would say that what a diagonal matrix represents is the fact that each of the $n$ variables required to specify a vector are decoupled. For example, in order to find the new $x_2$, one only needs to look at the old $x_2$, and do what the matrix says.
When we "diagonalize" a matrix, we're finding a way to describe each vector (that is, $n$ independent "pieces of information") that are similarly decoupled as far as the transformation is concerned. So, for example, the matrix
$$
A = \pmatrix{0&1\\4&0}
$$
takes a vector $x = (x_1,x_2)$ and produces a new vector $Ax = (x_2,2x_1)$. There's a nice symmetry to that; in particular, applying $A$ twice gives us the vector $A^2 x = (2x_1,2x_2)$, which is to say that $A$ acts like a diagonal matrix whenever you apply it an even number of times.
However, I would argue that we get a clearer picture of what $A$ does if we diagonalize it. In particular, if we write a vector as $x = a_1(1,2) + a_2(1,-2)$, $A$ gives us the new vector
$$
Ax = a_1 A(1,2) + a_2 A(2,1) = 2a_1(1,2) - 2a_1(1,-2)
$$
In particular, one we know the two pieces of information $a_1$ and $a_2$, we can figure out the new vector using these pieces separately, without having them interact.
So, we see that this antidiagonal $A$ is nice, but just not nearly as simple as the "diagonal version" of the transformation.
Symmetric matrices
Symmetric matrices are particularly nice when we care about dot-products. Dot products are needed whenever you want to think about the angle between vectors in some capacity.
In particular: if we define the dot-product
$$
(x_1,\dots,x_n) \cdot (y_1,\dots,y_n) = x_1y_1 + \cdots x_n y_n
$$
Then a symmetric $A$ will have the property that
$$
(Ax) \cdot y = x \cdot (Ay)
$$
Ultimately, this whole thing connects back to diagonal matrices since every symmetric matrix can be diagonalized in the sense described above. The fact that this can be done is known as the spectral theorem.
Persymmetric matrices, however, don't act in a particularly nice way with respect to any usual operations (like the dot product).
Imagine you want to project all $\mathbb{R}^3$ onto a plane $\pi$: necessarily some points from $\mathbb{R}^3$ have to go to same points in $\pi$ (simply speaking there are too many points in $\mathbb{R}^3$, you have to squeeze $\mathbb{R}^3$ to become a plane), so this map cannot be invertible. Defining this map with a matrix $A$ means that the matrix cannot be invertible because you cannot map every point from $\pi$ from it's respective point, where they came from, in $\mathbb{R}^3$. The only thing you can say about a projection matrix is that if $A$ is a projection matrix, then it's idempotent ($A^2=A$). In fact if you take the same map as before and try to apply it to the plane $\pi$ it has to get you again to $\pi$. The only invertible projection is the identity, which can be easily proven with the fact that a projection is idempotent.
Conversely if $A$ is a matrix of an isomorphism, by definition an isomorphism is a map one-to-one, so it maps every point in the domain to one and one only point into the codomain. So if you want to come back from the codomain to the domain you don't have any choice over taking a point from the codomain and making it come back from the point where it came from the first place. So an isomorphism is clearly invertible and then you can compute the inverse matrix $A^{-1}$ that gives you the inverse map
Best Answer
I will assume we are in $M_n(\mathbb{R})$, since you want your matrix to be orthogonal.
The spectrum of a skew-symmetric matrix is contained in $i\mathbb{R}$, so the spectrum of $A=I-S$ does not contain $0$ and $A$ is invertible, in $M_n(\mathbb{C})$ first, hence in $M_n(\mathbb{R})$.
The matrix we are interested in is $A^*A^{-1}$. Note that $A$ and $A^*$ commute.
Now $$ A^*A^{-1}(A^*A^{-1})^*=A^*AA^{-1}(A^{-1})^*=A^*AA^{-1}(A^*)^{-1}=A^*A(A^*A)^{-1}=I. $$
So $A^*A^{-1}$ is indeed orthogonal.