Let $\varDelta$ stand for your diagonal matrix to write the factorization as $S=O\varDelta O'$. Now, rewrite a bit more:
$$
S=(O\varDelta O^T)(OO')=\varSigma U,
$$
with $\varSigma$ symplectic positive definite and $U$ symplectic and orthogonal. Thus we are looking for a polar decomposition. In fact such a decomposition is unique, and given by
$$
\varSigma=(SS^T)^{1/2},\quad U=(SS^T)^{-1/2}S.
$$
Here we use that a positive semidefinite (symplectic) matrix has a unique positive definite (symplectic) square root. Now, once we have $\varSigma$, it can be diagonalized with a symplectic orthogonal linear change $O$ (again quite constructive), and we get the data we sought. For the "basic facts" one can look at Gosson's book, for instance.
No, not all symplectic $(0,1)$-matrices are block-triangular. The following matrix is symplectic:
$$\begin{pmatrix}0&1&0&0\\1&0&1&0\\0&1&0&1\\0&0&1&0\end{pmatrix}$$
A computer search found 44 counterexamples in dimension $4\times 4$, so here is another less symmetric one:
$$\begin{pmatrix}1&1&0&1\\1&1&1&0\\1&0&1&0\\0&1&0&1\end{pmatrix}$$
WolframAlpha code to check it's symplectic:
[[1,1,0,1],[1,1,1,0],[1,0,1,0],[0,1,0,1]]^T*[[0,0,1,0],[0,0,0,1],[-1,0,0,0],[0,-1,0,0]]*[[1,1,0,1],[1,1,1,0],[1,0,1,0],[0,1,0,1]]
Best Answer
As mentioned above, the groups are all isomorphic, however they are not all the same subset of $M_{2n}(\mathbb R)$.
Example. From MathWorld we have the following nice examples of symplectic matrices:
$$I_4, X = \begin{pmatrix}1&0&0&1\\0&1&1&0\\0&0&1&0\\ 0&0&0&1\end{pmatrix}, Y = \begin{pmatrix}0&1&0&1\\1&0&1&0\\0&0&0&1\\0&0&1&0\end{pmatrix}$$
Let's now try a different skew-symmetric nondegenerate matrix like $$J = \begin{pmatrix}0&0&2&0\\0&0&0&1\\-2&0&0&0\\0&-1&0&0\end{pmatrix}$$
A computation shows $X^TJX \neq J$. You can copy this following into WolframAlpha to check it:
Please note that one should not be that surprised by this! The same thing happens for instance with orthogonal groups. If one changes to a basis that is not orthonormal, the orthogonal matrices expressed in the new basis will no longer belong be in the standard subset $O_n \subset M_n(\mathbb R)$.
There is a little disparity between Wikipedia's definitions of symplectic matrix and orthogonal matrix in this sense. Orthogonal groups are of the form $A^TIA = I$ where $I$ is the identity matrix, but for a general nondegenerate symmetric bilinear form one replaces $I$ with any invertible symmetric matrix $S$. The group $\{A \in GL_n(\mathbb R) | A^TSA = S\} \cong O_n$ is certainly not equal to $O_n$ for all the same reasons as above.