Even if strictly speaking it's not true, it's always usefull to rememeber that morally $$K_i=q_i^{H_i}:=\exp(H_i\log(q_i))$$ where $q_i=q^{d_i}$ for some integer $d_i$ attached to the Cartan matrix of $\mathfrak g$, and where $H_i$ is the $i$th Cartan generator of $\mathfrak g$. This explain why the classical limit of $K_i$ should be 1.

Now set 
$$G_i=\frac{K_i-K_i^{-1}}{q_i-q_i^{-1}}$$

Then using again this heuristic it's easliy seen that you have
$$\lim_{q\rightarrow 1} G_i= \lim_{q\rightarrow 1}\frac{q_i^{H_i}-q_i^{- H_i}}{q_i-q_i^{-1}}= H_i$$

It's now rather clear how to prove that it indeed work, since $G_i$ is precisely the R.H.S. of the relation you want to modify, so just replace it by
$$[E_i,F_j]=\delta_{i,j} G_i$$
and add the relaiton
$$(q-q^{-1})G_i=K_i$$
which is nothing but the definition of $G_i$. That you get the same algebra is obvious because you did not really change anything. The former relation does not depend on $q$ anymore so gives you the appropriate relation at the classical limit, and the latter relation just become trivial.

Then following again the heuristic that $K_i$ should goes to 1 at the limit, take the quotient by the ideal generated by $K_i-1$, and you get the classical envelopping algebra.