Proving that two sets are diffeomorphic I have the following two sets 
$\mathcal{S}= \left\lbrace  (x,y,z,w) \in \mathbb{R}^4 \mid x^2+y^2- z^2-w^2 = 1 \right\rbrace$
and $\mathcal{S}' = \left\lbrace  (x,y,z,w) \in \mathbb{R}^4 \mid x^2+y^2- z^2-w^2 = r \right\rbrace$
for some non-zero real number $r$.
I need to show that these two sets are diffeomorphic.
I considered taking two cases for $r$ ($r>0$ and $r<0$);
then considering the map $f :\mathcal{S} \rightarrow \mathcal{S}' $ where $f( x,y,z,w) = \sqrt{|r|} ( x,y,z,w) $ when $r>0$.
Is this correct?
Thank you.

As you wrote yourself, your argument  works only for $r\gt 0$.  
If $r\lt 0$ the trick is to notice that the change of variables $x=Z,y=W,z=X, w=Y$ shows that $S$ is diffeomorphic to $Z^2+W^2-X^2-Y^2=1$ and thus to $x^2+y^2-z^2-w^2=-1$.
You can then apply your argument to show that you can replace the $-1$ on the right hand side of the equation $x^2+y^2-z^2-w^2=-1$ by any negative number $r$ to prove that $S$ is also diffeomorphic to the manifold $S'$ given by $x^2+y^2-z^2-w^2=r$ for any $r\lt 0$.
