Does Cauchy-Schwarz Inequality depend on positive definiteness?

Yes, Cauchy-Schwartz definitely depends on definiteness. It fails for indefinite forms: one can have $\left<u,u\right>=0$ but $\left<u,v\right>>0$. For instance consider $\left<(a,b),(c,d)\right>=ac-bd$ and $u=(1,1)$ and $v=(1,-1)$.

If $\|a\| = \|b\| = 0$, then \begin{align*} 0 & \leqslant \|a + b\|^2 = \|a\|^2 + \|b\|^2 + 2\langle a, b \rangle = +2\langle a, b \rangle, \\ 0 & \leqslant \|a - b\|^2 = \|a\|^2 + \|b\|^2 - 2\langle a, b \rangle = -2\langle a, b \rangle, \end{align*} therefore $\lvert\langle a, b \rangle\rvert = 0 \leqslant \|a\|\|b\|$.

Suppose, on the other hand, that $\|b\| > 0$. Then a fairly standard argument applies. Define $\lambda = \langle a, b \rangle/\|b\|^2$. Then $\langle a - \lambda b, b \rangle = \langle a, b \rangle - \lambda \|b\|^2 = 0$, therefore \begin{align*} 0 & \leqslant \langle a - \lambda b, a - \lambda b \rangle \\ & = \langle a, a - \lambda b \rangle - \lambda\langle b, a - \lambda b \rangle \\ & = \langle a, a - \lambda b \rangle - \lambda\langle a - \lambda b, b \rangle \\ & = \langle a, a - \lambda b \rangle \\ & = \|a\|^2 - \lambda \langle a, b \rangle, \end{align*} therefore $$ \langle a, b \rangle^2 = \lambda\|b\|^2\langle a, b \rangle \leqslant \|a\|^2\|b\|^2, $$ therefore $$ \lvert\langle a, b \rangle\rvert \leqslant \|a\|\|b\|. $$ The argument is similar when $\|a\| > 0$. So the Cauchy-Schwarz inequality holds in all cases.

Addendum

It appears (see my series of shame-faced comments below for details) that this argument is merely an obfuscation of what is surely the most "standard" of all proofs of the Cauchy-Schwarz inequality. It is the one that is essentially due to Schwarz himself, and he had good reasons for using it, quite probably including the fact that it makes no use of the postulate that $\langle x, x \rangle = 0 \implies x = 0$! In a modern abstract formulation, it goes as follows (assuming, of course, that I haven't messed it up again). For all real $\lambda$, we have $\|u\|^2 - 2\lambda\langle u, v \rangle + \lambda^2\|v\|^2 = \|u - \lambda v\|^2 \geqslant 0$. Therefore, the discriminant of this quadratic function of $\lambda$ must be $\leqslant 0$. That is, $\langle u, v \rangle^2 \leqslant \|u\|^2\|v\|^2$; equivalently, $\lvert\langle u, v \rangle\rvert \leqslant \|u\|\|v\|$.