So the statement you made is true, but does not prove what you want it to prove.

[roam]

If there are integers s,t with as+bt=6, this implies that gcd(a,b)=6, right?

And if gcd(a,b)=6, does this necessarily mean that a and b are not relatively prime since their gcd is not 1? (I have read that two integers a and b are relatively prime if gcd(a,b)=1).

 

[rasmhop]

If there are integers s,t with as+bt=6, this implies that gcd(a,b)=6, right?

No. It simply implies that gcd(a,b) divides 6. For instance take a=4, b =6, then gcd(a,b) = 2 but:
$$0a+1b = 6$$

And if gcd(a,b)=6, does this necessarily mean that a and b are not relatively prime since their gcd is not 1? (I have read that two integers a and b are relatively prime if gcd(a,b)=1).

Yes, 6 divides them both so they have a common factor besides 1 (2,3,6 are all common factors). That gcd(a,b)=1 is actually a pretty common definition of integers being relatively prime.

By the way if you wanted to use this argument to prove that a and b are not relatively prime, then unfortunately that doesn't work. Consider for instance:
a = 2, b= 3
which are definitely relatively prime as they are both prime, but:
$$6b+(-6)a = 6$$