Yes they can. This follows from singular value decomposition. Let S be the matrix representation of a shear transformation. There exist rotation matrices R, B and a diagonal matrix D such that S = RDC, where C is the transpose of B. D is the matrix representation of a scaling transformation and R, B are the matrix representations of rotation transformations. Since S is a product of rotation and scaling matrices, its corresponding linear transformation is a composition of rotations and scalings.
It would ordinarily be weird to represent shear transformations using rotations and scalings because shear matrices are elementary. But it checks out.
OK, point taken. I considered "scaling" in a less general sense (scalar multiple of the unit matrix), while you want to allow arbitrary diagonal entries. My definition is to my knowledge the common one in linear algebra textbooks because in yours, the feasible maps depend on the chosen basis.
EDIT: To state my point more clearly: in textbooks, "scaling" is the linear map that is induced by the "scalar multiplication" in the definition of the vector space (that is why both terms start with "scal").
It would ordinarily be weird to represent shear transformations using rotations and scalings because shear matrices are elementary. But it checks out.