Yes, precisely.

I agree it is not obvious a priori. But my answer explains that any function $F(z,s)$ satisfying just (2), (3) and (5) (even a weaker version of (5) just asking for moderate growth) will satisfy $B(s)/A(s)=\xi(2-2s)/\xi(2s)$ when $Re(s)>1$, since for fixed $s$ in that range it will be a multiple of $E(z,s)$. So in fact one knows a lot more that $A(s)=B(1-s)$.

Well, often the normalisation used gives $A(s)=1$. In any case, one always has $B(s)/A(s)=\xi(2-2s)/\xi(2s)$; see e.g. Bump's book. So at least taking $A(s)=\xi(2s)$ 'clears some denominators', and is necessary if you want to avoid having poles of the Eisenstein series at zeros of $\xi(2s)$.

I see. But any function satisfying (2) and (3) has a constant term which will be of the form $Ay^s + B y^{1-s}$ (by invariance of the Laplacian under translations). For $Re(s)>1$, you can't have $A=0$ for the reasons given in my answer. So you are just rescaling to $A=\xi(2s)$.

Dear Hugo, I am a bit confused since I did not mention the constant term, just a growth estimate. If you can be a bit more precise about what you would consider more conceptual, then I'll try to think about it!