Cosets modulo $(2+i)$ in $\mathbb{Z}[i]$

Your approach is a good one. (There are others, as in the linked material at the end of this answer, but there is no reason not to continue pursuing your approach.)

Why don't you try to show that $A = I$?

Given $x$ and $y$ (integers) such that $5 | 2x + y$, try to find $a$ and $b$ so that $x = 2 a - b$ and $y = a + 2b$. (You can just try to solve for $a$ and $b$; when you do, your divisibility condition will come in useful to ensure that $a$ and $b$ are integers.)

Once you do this, you'll be finished, since, as you observed, $A$ has index $5$ in $\mathbb Z[i]$.

For a different perspective on this kind of question, you could look at this answer (and the other answers posted there). But I would encourage you to push your original approach all the way through!

Exercise: Show, in general, that the number of Gaussian integer equivalence classes, modulo a Gaussian integer $z$ is exactly $N(z)$, where $N$ is the norm function.