Let me add my own approach, thereby offering a context and motivation for the identity at hand. Of course, it is not the easiest technique but that was not my goal or intent in asking the question above.

As I mentioned, things came out of determinantal consideration.

The story began with this MO question, in which I provided an alternative proof to the (well-known) special case of the $(n+1)\times(n+1)$ Hankel matrix
$$\det\left[\binom{2i+2j}{i+j}\right]_{i,j=0}^n=2^n.\tag 1$$
**Method 1.** Generalize (1) and prove that (check my argument here)
$$\det\left[\binom{2i+2j+2a}{i+j+a}\right]_{i,j=0}^n
=\prod_{j=0}^n\binom{2a+2j}{a+j}\binom{a+n+j}{2j}^{-1}.$$
In particular, setting $a=0$, the determinant (1) is computed by
$$\det\left[\binom{2i+2j}{i+j}\right]_{i,j=0}^n
=\prod_{j=1}^n\binom{2j}j\binom{n+j}{2j}^{-1}.\tag2$$
**Method 2.** Using the matrix decomposition (a product of triangular and diagonal matrices)
$$\left[\binom{2i+2j}{i+j}\right]_{i,j=0}^n
=\left[\binom{2i}{i-j}\right]\cdot D\cdot\left[\binom{2i}{i-j}\right]^T;\tag 3$$
where $D$ is the diagonal matrix $D=\text{diag}(1,2,2,\dots,2)$. Now, direct calculation in (3) generates
$$\det\left[\binom{2i+2j}{i+j}\right]_{i,j=0}^n=2^n. \tag4$$

Finally, equating (2) and (4) leads to the identity in our problem.