Why is the point at infinity on Edwards Curve different to Weierstrass curves? If I understand correctly, the identity point on all elliptic curves is the point at infinity.
But on the Edwards curve, this can be written in Affine form?

Does this have something to do with the fact that Edwards curve formulas are complete?

What would the point (0,1) be on the Weierstrass curve signify? If b=1, then we can represent this point on the Weierstrass curve?

It still has not clicked with me, the fact that the point at infinity is in the group, yet it cannot be represented? But somehow it can b represented on the Projected form, which is another form of the same curve? Whereas Montgomery is a different curve entirely?
In mathematics, every group need and identity element. The points on the elliptic curves forms a group. Therefore they need an identity element, too.
The identity element is determined by the addition law of the curve. Some curves need a point at infinity $\mathcal{O}$ as the identity element (neutral element).
In the Edwards curve, the neutral element is chosen as $(0,1)$ and this can be represented in the affine coordinates. Other coordinate like $(0,-1)$ can be identity, however, this requires different formulas.

If I understand correctly, the identity point on all elliptic curves is the point at infinity.

No not necessarily. Edwards curves doesn't need a point at infinity.

But on the Edwards curve, this can be written in Affine form?

Yes, it is $(0,1)$

Does this have something to do with the fact that Edwards curve formulas are complete?

No, it is just by the equation of the curve and the defined addition law. This is very similar to the analogous to clock

What would the point (0,1) be on the Weierstrass curve signify? If b=1, then we can represent this point on the Weierstrass curve?

Converting formulas exist like this one

Fast Algorithm for Converting Ordinary Elliptic Curves into Binary
Edward Form


it still has not clicked with me, the fact that the point at infinity is in the group, yet it cannot be represented? But somehow it can b represented on the Projected form, which is another form of the same curve? Whereas Montgomery is a different curve entirely?

An birationally equivalence doesn't say about the representation of the elements. Consider the isomorphism of two group, they are same but they can be completely defined over different sets.
The the neutral element can be defined but you force it to be defined over the affine coordinates. Some curves need the protective coordinates so that every element can be represented with some coordinates.