All that b does to f(x) is translates it vertically, so lets start by taking a look at g(x) = ax+9/x.

The span of g(x)'s range must be less than 4 for there to be some b to translate g(x) into f(x) such that -2<f(x)<2 over the entire domain of [1,9].

 If the range has a span of at least four then there will not exist any such translation

So then, what is the possible range span of g(x)?  There are three possible extremes: g(1), g(9), or a local extrema of g(x) inside the interval [1,9].  

The local extrema occurs when g'(x)=0, then g'(x)=a-9/x^2=0 has a solution when x=sqrt(9/a).  This extema will occur in the interval [1,9] when 1/9<a<1.  

Looking at g"(x)=18/x^3 being positive everywhere in [1,9] we can conclude that g(sqrt(9/a)) is a local minimum. 

So three cases arise: a<1/9, 9<a, and 1/9<=a<=9.  In the first two cases only the endpoints g(1) and g(9) matter.  Only in the third case do we need to consider g(sqrt(9/a)).

Case 1: a<=1/9

g(1)-g(9) = -8a+8.  -8a+8>4 when a<1/2.  This covers the entire case of a<=1/9.  So the span of the range of g(x) is always greater than 4 in this case.

Case 2: 9<=a

g(9)-g(1) = 8a-8.  8a-8>4 when a>3/2.  This covers the entire case of 9<=a.  So the span of the range of g(x) is always greater than 4 in this case.

Case 3: 1/9<=a<=9

g(9) or g(1) may be greater depending on the value of a; so comparing g(9)>=g(1): 9a+1>=a+9 is true when a>=1; thus g(9) is the greater endpoint when a>=1, and g(1) otherwise.  

So there two subcases and one edge case: 1/9<=a<1, 1<a<=9, and edge case a=1.

Subcase 1: 1/9<=a<1

g(1) and g((sqrt(9/a)) are the extrema.  Then g(1)-g((sqrt(9/a)) = a+9-6sqrt(a) = (3-sqrt(a))^2.  This is greater than 4 for any a<1.  So the the span of the range of g(x) is always greater than 4 in this case.

Subcase 2: 1<a<=9

g(9) and g((sqrt(9/a)) are the extrema.  Then g(9)-g((sqrt(9/a)) = 9a+1-6sqrt(a) = (3*sqrt(a)-1)^2.  This is greater than 4 for any a>1.  So the the span of the range of g(x) is always greater than 4 in this case.

At this point we have shown for any a!=1 then the span of the range of g(x) and f(x) must be larger than 4 and then f(x) must necessarily have part of the function outside -2<f(x)<2.

Edge case a=1.

g(1) = g(9) = 10 and g((sqrt(9/1)) = g(3) = 6.  This span is exactly 4.  Then b=-8 will make it so f(x) exactly fits within -2<=f(x)<=2.  Any other value of b will cause a portion of the range to fall outside of -2<=f(x)<=2

So the specific f(x) = a-8+9/x will be the only f(x) such that some continuous range is not outside of the bounds of -2<=f(x)<=2.  However it still has three points (1,2), (3,-2) and (9,2) on the boundary.

Thus the claim is true that for any (a,b) there must be some x0 for which |f(x0)|>=2.

If the claim was slightly tweaked into a strict inequality |f(x0)|>2, then there would be a single f(x) to fail that tweaked claim, namely f(x) = a-8+9/x.