The following identity is well-known and there are a few proofs to it (see Bijective proof problems, by R P Stanley, for this and similar formulae): $$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}=4^n \qquad \iff \qquad \sum_{j+i=n}\binom{2j}j\binom{2i}i\frac1{4^n}=1.\tag1$$
QUESTION. Is there a $q$-analogue to this "innocent-looking" identity in equation (1)?
We start with $q$-binomial theorem $$ (x+y)(x+qy)\cdots(x+q^{n-1}y)=\sum q^{k\choose 2}{n\choose k}_qy^k x^{n-k}.\quad\quad\quad(\heartsuit) $$ Put $n=a+b$ in $(\heartsuit)$ and consider separately the first $a$ multiples in LHS and the last $b$ multiples. We get $$ \left(\sum q^{j\choose 2}{a\choose j}_qy^j x^{a-j}\right)\cdot \left(\sum q^{{i\choose 2}+ai}{b\choose i}_qy^i x^{b-j}\right)= \sum q^{k\choose 2}{a+b\choose k}_qy^k x^{a+b-k}. $$ Taking coefficients of $x^ky^{n-k}$ we get $$ \sum_{j+i=k} q^{{j\choose 2}+{i\choose 2}+ai}{a\choose j}_q {b\choose i}_q=q^{k\choose 2}{a+b\choose k}_q.\quad\quad\quad\quad\quad\quad\quad\quad(\clubsuit) $$ For fixed $k$ (and $q$) both LHS and RHS of $(\clubsuit)$ are polynomials in $q^a$ and $q^b$. Thus we may substitute $q^a=q^b=q^{-1/2}$ (I do this because in the ordinary version the identity is equivalent to the $a=b=-1/2$ version of the Vandermonde--Chu convolution $\sum_{a+b=k} {a\choose i}{b\choose j}={a+b\choose k}$). We have $${-1\choose k}_q=\frac{(q^{-1}-1)(q^{-2}-1)\cdots(q^{-k}-1)}{(q^k-1)\cdots(q-1)}=q^{-{k+1\choose 2}}(-1)^k.$$ Next, denote $q=\tau^2$, then $${-1/2\choose j}_q=\frac{(q^{-1/2}-1)(q^{-3/2}-1)\cdots(q^{-(2j-1)/2}-1)}{(q^j-1)\cdots(q-1)}\\=\tau^{-j^2}(-1)^j\frac{(\tau-1)(\tau^3-1)\cdots (\tau^{2j-1}-1)}{(\tau^2-1)(\tau^4-1)\cdots (\tau^{2j}-1)}\\=\tau^{-j^2}(-1)^j{2j\choose j}_{\tau}\frac1{(1+\tau)^2(1+\tau^2)^2\cdots (1+\tau^j)^2}.$$ So $(\clubsuit)$ reads as $$ \sum_{j+i=k} \tau^{j} {2j\choose j}_{\tau}{2i\choose i}_\tau\frac1{(1+\tau)^2(1+\tau^2)^2\cdots (1+\tau^j)^2\cdot (1+\tau)^2(1+\tau^2)^2\cdots (1+\tau^i)^2}=1. $$
This is an analogue of the identity you ask about in the sense that for $\tau=1$ we get it.
Note by OP: If we write the $j$-factorial analogue as $(q)_j=\prod_{k=1}^j\frac{1-q^k}{1-q}$, then the last identity takes the form $$ \sum_{j+i=n} \tau^{j} {2j\choose j}_{\tau}{2i\choose i}_\tau \frac{(\tau)_j^2(\tau)_i^2}{(\tau^2)_j^2(\tau^2)_i^2}=1. $$
