As dkaeae indicated in his comment, a right-moving or staying Turing machine (TM) is essentially a finite deterministic automaton (FDA). Here's a proof.
Let $M$ be such a machine, whose transition rules is in the form of $\delta(q, \gamma)=(t, \beta, d)$ where $q$ is the current state, $\gamma$ is the contents of the current cell, $t$ is the new state, $\beta$ is the symbol that replaces $\gamma$ in the current cell, and $d$ is the direction to move the head, which is either $R$ or $S$, meaning moving right or staying put respectively.
Let us check the behavior of $M$ when it has just applied a transition rule of the form $\delta(q, \gamma)=(t, \beta, S)$. Now $M$ is in state $t$ on top of $\beta$. In the next step, $M$ will read the symbol $\beta$, applying the unique rule on the pair $(t,\beta)$.
Suppose that rule is $\delta(t, \beta)=(u, \alpha, d)$.
Then $M$ will change state to $u$, rewrite the current cell to $\alpha$ and move in the direction of $d$.
We have found that once $M$ was in state $q$ on top of $\gamma$, it will change state to $u$, rewrite the current cell to $\alpha$ and move in the direction of $d$.
So we can replace the transition rule $\delta(q, \gamma)=(p, \beta, S)$ by $\delta(q, \gamma)=(u, \alpha, d)$ in the specification of $M$ without changing the language accepted by $M$.
Suppose that rule is not defined.
Then $M$ will halt. We can remove the transition rule $\delta(q, \gamma)=(p, \beta, S)$ in the specification of $M$ without changing the language accepted by $M$.
Applying the replacement or removal above repeatedly until there is no rule in $M$ that tells $M$ to stay in the same cell, we find that $L(M)$ is the language accepted by a right-moving only TM.
This question and answer tells us a right-moving only TM is essentially a DFA. Basically, a rule in a right-moving only TM, $\delta(q, \gamma)=(t, \beta, R)$ corresponds to $\delta(q, \gamma)=(t)$, a rule in the corresponding DFA. Since the language of a DFA is decidable, so is $L(M)$.