That does not show what was asked. The original questions was to show that any second order differential equation could be put in that form. Chris L has only showed that you can write an equation of the form in that form. What if b(x) is not equal to a'(x)?
Just as with first order equations, try to find an "integrating factor", u(x). Look at (a(x)u(x)y')'= a(x)u(x)y"+ (a'(x)u(x)+ a(x)u'(x))y' and compare that with a(x)u(x)y"+ b(x)u(x)y'. In order that those be the same we must have a'(x)u(x)+ a(x)u'(x)= b(x)u(x) which is the same as the first order, separable, equation a(x)u'(x)= (b(x)- a'(x))u(x). Integrating gives the integrating factor, u(x). Multiplying both sides of the equation by that u(x) puts it in the form required.