Yes, you can find a and b provided you have u(n+1), u(n) and u(n+1), for example.
We have:
u(n)=au(n-1)+b
u(n+1)=au(n)+b
So:
u(n)-u(n+1)=a(u(n-1)-u(n)),
a=(u(n)-u(n+1))/(u(n-1)-u(n)).
b=u(n)-au(n-1),
b=u(n)-u(n-1)(u(n)-u(n+1))/(u(n-1)-u(n)),
b=(u(n+1)u(n-1)-(u(n))²)/(u(n-1)-u(n)).
You can use any two pairs of consecutive terms to form a system of two equations to find a and b.