Class of languages recognised by a Forgetful Turing Machine
The class languages recognized by FTMs are the class of regular languages. A FTM can completely simulate a DFA, so each regular language can be recognized by a FTM. On the other hand, if we can prove each FTM recognizes a regular language, we have proved the assertion.

Given a FTM $F$ with the set of states $Q$, we use the Myhill–Nerode theorem to show the language it recognizes is regular. Given a string $x$, we ask the following query:


If $F$ runs on $x$, does $F$ ever leave $x$ to the right? If so, in which state is $F$ when it first moves to the first position to the right of $x$? If not, does $F$ accept $x$?


... and the following $|Q|$ queries for each state $q\in Q$:


If $F$ runs on $\underbrace{a\cdots a}_\text{the number of $a$ = $|x|$}$, but is started on the rightmost $a$ and in state $q$, does $F$ ever leave the input to the right? If so, in which state is $F$ when it first moves to first position to the right of the input? If not, does $F$ accept the input finally?


We collect the strings that give the same answers to the $|Q|+1$ queries above into the same class. Since there are $(|Q|+2)^{|Q|+1}$ possible answers to the $|Q|+1$ queries above, there are only finite many classes. Next, we are going to prove that for any strings $x,y$ in the same class, i.e., that give the same answers to the $|Q|+1$ queries, for any string $z$, $F$ accepts either both or neither of $xz$ and $yz$, which completes the proof by the Myhill–Nerode theorem.

Consider that $F$ runs respectively on $xz$ and $yz$. Note the answers to query 1 are the same for $x$ and $y$, there are two cases:


$F$ never leaves $x$ to the right on $xz$, and $F$ never leaves $y$ to the right on $yz$. In this case, $F$ runs as if the inputs are respectively $x$ and $y$. Since the answers to query 1 are the same for $x$ and $y$, $F$ accepts either both or neither of $xz$ and $yz$.
$F$ ever leaves $x$ to the right on $xz$, and $F$ ever leaves $y$ to the right on $yz$. In this case, $F$ is in the same state when it first moves to the leftmost position of $z$, so $F$ behaves the same until it returns respectively to the rightmost position of $x$ and $y$. According to the definition of FTM, the input must become $a\cdots a$ when $F$ returns, and we can apply a recursive analysis according to query 2 (you can use mathematical induction to write a strict proof). As a result, $F$ accepts either both or neither of $xz$ and $yz$.


Of course, the analysis above assumes the tape of the FTM is restricted such that the head cannot leave the input to the left. The proof is similar without this assumption.