Sometimes you can
Actually, SEJPM has it mostly right; given a triplet $(cG, eG, aG)$, there will always be a possible $B, D$ such that $aG + B = cG$ and $aG + D = eG$.
However, they does leave open the question, given a triplet $(C, E, A)$, is it representable as $(cG, eG, aG)$? If not, the first 5 equations cannot hold, and so we can know that the $C, E, A$ values we were given were not formed as specified.
A simple analogy in $\mathbb{Z}$ (the integers) would be if $G=2$, and we were given the triple $(4, 2, 7)$. Even though there are values $B, D$ that are consistent with that triplet, we can also see that there are no sets of values $a, b, d$ that make $A$ odd (as $aG$ is always even; remember, $a$ cannot be a fraction), and so we can reject that.
So, the obvious question is: can that sort of thing happen with elliptic curves? Well, the answer to that is "perhaps" (depending on the elliptic curve and whether you know the value $G$).
There are elliptic curves that don't allow you to reject any triplet (as long as you know that $G$ is not the 'identity element')
There are elliptic curves that, if you know what $G$ is, may allow you to reject some triplets (depending on what the value $G$ is).
There are elliptic curves, that, even if you don't know what $G$ is, will allow you to reject some.
However, unless the elliptic curve was especially crafted, a random triplet has a good possibility of being accepted by any of the above tests.