Generating function proof of floor identity

Nice temptation. Let us consider the ordinary generating function of the $\{\frac{c_n}{n!}\}_{n\geq 0}$ sequence, which is also the exponential generating function of the $\{c_n\}_{n\geq 0}$ sequence:

$$ f(x)=\sum_{n\geq 0}\frac{x^n}{n!}\binom{n}{\lfloor n/2\rfloor} = \sum_{n\geq 0}\frac{x^n}{\lfloor n/2\rfloor!\lceil n/2\rceil!}=\sum_{m\geq 0}\frac{x^{2m}}{m!^2}+\frac{x^{2m+1}}{m!(m+1)!}=I_0(2x)+I_1(2x).$$ Now

$$\sum_{k=0}^{n}\binom{n}{k}c_k c_{n-k}=n!\sum_{k=0}^{n}\frac{c_k}{k!}\cdot\frac{c_{n-k}}{(n-k)!}=n![x^n]f(x)^2 $$ so the claim is equivalent to $$ n![x^n]\left(I_0(2x)+I_1(2x)\right)^2 = \binom{n}{\lfloor n/2\rfloor}\binom{n+1}{\lfloor (n+1)/2\rfloor} $$ or to

$$ (I_0(2x)+I_1(2x))^2 = \sum_{m\geq 0}\frac{x^{2m}}{(2m)!}\binom{2m}{m}\binom{2m+1}{m}+\frac{x^{2m+1}}{(2m+1)!}\binom{2m+1}{m}\binom{2m+2}{m+1}. $$ Now I can show that both sides are entire functions with the same Laplace transform, which is related to elliptic integrals. But I guess there are simpler ways, currently eluding me.

All right, here they are: $$[x^n]\left(\sum_{m\geq 0}\frac{x^m}{m!^2}\right)^2 = \sum_{k=0}^{n}\frac{1}{k!^2(n-k)!^2}=\frac{1}{n!^2}\sum_{k=0}^{n}\binom{n}{k}^2=\frac{\binom{2n}{n}}{n!^2}=\frac{(2n)!}{n!^4}$$ so the whole exercise can be seen as a consequence of the Chu-Vandermonde identity.