If I understand your question correctly, you are essentially asking if points in Edwards and Montgomery curves can be represented in Weierstrass coordinates. This is true; in fact, any elliptic curve over a field of characteristic $\neq2,3$ can be represented in Weierstrass form $\mathcal{E}_{w}^{a, b} : y^2 = x^3 + ax + b$, and by extension its points can too.
The question, then, becomes “how easy is the conversion”?
For Montgomery curves of the form $\mathcal{E}_m^{A, B}: By^2 = x^3 + Ax^2 + x$, it is surprisingly straightforward. You can map the point $(x, y) \in \mathcal{E}_m^A$ to the Weierstrass curve $\mathcal{E}_w^{1/B^2-A^2/3B^2,A(2A^2 - 9)/27B^3}$ via the map $((x + A/3)/B, y/B)$. The inverse map is obvious, $(Bx - A/3, By)$. If $B=1$, which is the case for every Montgomery curve used in practice, this simplifies substantially.
For twisted Edwards curves $\mathcal{E}_{e}^{a, d} : ax^2 + y^2 = 1 + dx^2y^2$ (where Edwards curves are the subset where $a = 1$) there also is a simple map to a Weierstrass curve, which consists of converting the curve to Montgomery format first. In particular, we can map $\mathcal{E}_e^{a, d}$ to $\mathcal{E}_m^{2(a + d)/(a - d), 4/(a - d)}$ via $(x, y) \mapsto \left(\frac{1 + y}{1 - y}, \frac{1 + y}{x - xy}\right)$, with inverse $(x, y) \mapsto \left(\frac{x}{y}, \frac{x-1}{x+1}\right)$. Composing theses maps with the above, we get maps to the appropriate Weierstrass curve $\mathcal{E}_w^{a, b}$. None of these conversions are terribly expensive compared to the cost of a scalar multiplication.
