This is a number theory problem actually. Anyways we show the maximum is $2$. Clearly, suppose $(a_0,b_0)$ are two solutions of the Pell's equation $4x^2-y^2=3$. Now $a=8(a_0^2-1),b=2(a_0^2-1)$ gives $ab=(4(a_0^2-1))^2, (a+2)(b+2)=(8a_0^2-6)(2a_0^2)=(2a_0b_0)^2$. So $2$ is achieved. 6 or 5 is impossible: Let $a(a+2),b(b+2)$ none of them are squares since they are one less than a square. Thus we cannot have them being a natural square. 4 is impossible: $ab=m^2,a(b+2)=n^2,(a+2)b=p^2,(a+2)(b+2)=q^2\implies mq=pn$ also see that $p^2+n^2+4=m^2+q^2\implies (m-n)^2=(p-q)^2+4(\text{since }mq=pn)\implies p=q$. But then $a=b$ contradicting distinctness. 3 is impossible: For this to happen one of these pairs must be simultaneously squares: $\{(ab,a(b+2)),(ab,(a+2)b),(a+2)(b+2),a(b+2)),((a+2)(b+2),(a+2)b)\}$ but multiply any two of them and since their product is a perfect square then $a^2+2a$ or $b^2+2b$ must be a perfect square as well. Which is impossible. Hence $\text{max}(S)\le 2$. $\blacksquare$

The only solutions to $ 4x^2-y^2=3 $ are $(x,y) = (1,1)$, so that your values of $a,b$ are $0$. However, replacing your equation with $\boxed{16x^2 - y^2 = 15}$, which has roots $(a_0, b_0) = (2,7)$ we can set $a=32(a_0^2 - 1), b=2(a_0^2 - 1)$, assuring us that $S=2$ works, then continue as before.

It is clear that $ a (a+2) = (a+1)^2 -1 , b (b+2) = (b+1)^2 -1 $ are not perfect squares . Note that if $ ab $ and $ a (b+2) $ were both perfect squares, then $ (ab) * a (b+2) $ must be a perfect square , which in turn means that $ b (b +2) $ must be a perfect square, which is not possible. Similarly at most one number of each of the pairs $(ab, b (a+2) ), ((a+2)(b+2), a (b+2)) , ((a+2)(b+2) , b (a+2)) $ can be a perfect square. Joining the bits of information, we conclude that at most two perfect squares can be found. Note that for $ (a,b) = (25,1) $ , we have $ ab = 5^2 , (a+2)(b+2) = 9^2 $, so $ 2$ is achievable. Therefore the maximum value of $S $ is $2$.