On the first attempt compare two groups of $9$ coins each: $G_1 = 1,3,5,7,9,11,13,15,17$ and $G_2 = 2,4,6,8,10,12,14,16,18$. Note that the $9$ grams coin $A$ and the $11$ grams coin $B$ cannot be on the same pan as their positions are consecutive, hence of different parity.

If there is equilibrium we claim that, moreover, neither $A$ nor $B$ is on the pans, i.e. they are at positions $19$ and $20$. Indeed suppose that $A$ or $B$ is on one of the pans. Equilibrium is impossible with exactly one exceptional coin; the other one must be on a pan too. Moreover for equilibrium both of them must be on the same pan; however we remarked that this is not so. Thus $A$ and $B$ are at positions $19$ and $20$, and since $B$ is the right neighbor of $A$, we find that the $11$ grams coin is the last one. Note that the case of equilibrium needs no further attempts.

Let $G_1$ be lighter than $G_2$. Then $A \in G_1$. Indeed if $A \in G_1$ then $G_1$ has only $10$ grams (it cannot contain $B$). Hence $G_1$ is lighter only if $B \in G_2$. So $B$ is at even position $2, 4, ..., 18$. But then $A$ is at the previous odd position $1, 3, ..., 17$, i.e. it is in $G_1$ contrary to the assumption.

Similarly let $G_1$ be heavier than $G_2$. Then $A \in G_2$. Otherwise $G_2$ has only $10$ grams coins and it is lighter only if $B \in G_1$. So $B$ is at an odd position $3, 5, ..., 17$; note is not $1$ as $B$ is preceded by $A$. But then $A$ is at the previous even position $2, 4, ..., 16$, i.e. $A \in G_2$, contrary to the assumption.

So in the case of non-equilibrium the first attempt finds a group of $9$ coin with $8$ of them having the same weight and the last one lighter. It is known how to find the lighter coin with $2$ attempts. Divide the coins into $3$ groups of $3$ and compare two groups. Regardless of the outcome this determines a group of $3$ coins containing the lighter one. It remains to compare two coins from $G$. Regardless of the outcome the lighter coin will be identified.