Joerg> One can regularize to get that one
(via abs( qexp(I*x) ) == 1 for some nice qexp()). The paper Jan L.\ Cie\'{s}li\'{n}ski: {Improved $q$-exponential and $q$-trigonometric functions}, arXiv:1006.5652v1 [math.CA], (29-June-2010). URL: \url{http://arxiv.org/abs/1006.5652}.} does it for the version where limit_{q-->1} gives the usual functions (exp(), cos(), sin()).
rwg> Caution to the casual: In this paper, superscripts of e, E and Script-E are *not* exponents. In fact, I don't even see a relation of the form Script-Exp_q1(x)^2 ~ Script-Exp_q2(2x), so I don't see much motivation, beyond nice q-difference and nice Pythagoras. Can we have q, Pythagoras, *and* identities?
Ohh, sort of. Using Cie ́slin ́ski's regularized q-exponential rexpq[q_, x_] -> QPochhammer[-1/2 (1 - q) x, q]/QPochhammer[1/2 (1 - q) x, q] and your bisected product observation (below), i.e., QPochhammer[q x, q^2]^2/QPochhammer[-q x, q^2]^2 == ( QPochhammer[-x, -q] QPochhammer[x, q])/(QPochhammer[-x, q] QPochhammer[x, -q]) we have rexpq[q^2, q x]^2 == rexpq[-q, (-1 + q) x] rexpq[q, (1 + q) x] which, as q->1, becomes (e^x)^2 = ?(-1) * e^(2x), where ?(-1) must be some flavor of 1. (Cie ́slin ́ski doesn't explain negative q.) This could take some getting used to.
[...] Joerg> The equation in question is
E(+q^2, -q*x)^2 = E(+q, -x) * E(-q, +x)
I stopped investigating, however, after looking into Whittaker/Watson chapter.21 and suspecting my finding is really a specialization of stuff given there.
It's just a bisected product but it might be a key to finding rexpq and regularized q-trig identities . All we do is replace some occurrences of 2 with 1+q, and some occurrences of 0 with 1-q !-) --rwg