Is it because glyceraldehyde -3-phosphate (a molecule which when enter glycolysis help produce ATP through substrate level phosphorylation) can be prepared without losing an ATP through this process?
No. Your supposition is incorrect — the phosphate in glyceraldehyde 3-phosphate has to come from somewhere, and it comes from glucose 6-phosphate. The reason a second ATP is required before you get to the triose phosphate stage in glycolysis is that you are generating two molecules of triose phosphate. In the pentose phosphate pathway (energy-producing non-oxidative branch) you are not generating two molecules of triose-P from one hexose-P, you are generating two hexose-P and one trisose-P from three hexose-P, as shown in my Diagrams 2 and 3 below. (The other three carbon atoms are lost as CO2.)
A good way of approaching this question is to ask “What can be produced by the pentose phosphate pathway that cannot be produced by glycolysis?”.
Glycolysis may be summarized in this context as:
Glucose + NAD+ ➝ Pyruvate + NADH + 2ATP
But to allow glycolysis to continue the NADH is reoxidized to NAD+:
Pyruvate + NADH ➝ Lactate + NAD+
So the sole product of glycolysis for the erythrocyte is ATP (primarily for active cation transport to maintain cell shape) lactate passes into the blood for recycling.
The Pentose Phosphate Pathway (the oxidative stage) is shown in Diagram 1, above, and may be summarized in the way done for glycolysis as:
Glucose + ATP + 2NADP+ ➝ Ribulose 5-P + CO2 + 2NADPH
Ribulose 5-P has two possible fates, but only one differs from glycolysis, so the two distinguishing products of the pathway are ribose and NADPH.
Ribose is important for nucleic acid synthesis, particularly in dividing cells, explaining the increased activity of the pentose phosphate pathway in those cells. This cannot be the reason for the high activity of the pathway in erythrocytes as they have no nuclei and do not divide. In fact the ribulose 5-P is fed back into glycolysis for generation of ATP as shown in diagrams 2 and 3.
NADPH, then, is the answer in this case. It is the reducing agent used in the cytoplasm for synthetic processes (unlike NADH used particularly for ATP generation in the mitochondria), and so the pentose phosphate pathway is found in cells such as adipose, mammary gland and liver, that synthesize fatty acids, and cells that synthesize steroids. However that is not there is no fatty acid or steroid synthesis in the erythrocyte.
NADPH is important in erythrocytes as it is the specific source of reducing power required to keep the molecule glutathione in a reduced state. This has an important protective role in reducing cellular molecules that have become oxidized by molecular oxygen, a problem that is more acute in erythrocytes than in perhaps any other cells as they are the carriers of molecular oxygen. The cell membrane is particularly prone to oxidative damage. You can read about this online in this chapter in Berg et al.