Equivalence of dfferent TM definitions
Yes they are equivalent and it is always possible. Informally: you can exchange any $S$ move state with pair of states that goes $L$ then $R$ (or in the other direction).

Formally:
Prove that standard Turing Machine is equivalent to class of Turing Machines with $S$ (no move).
Let $M$ denote standard TM, there exists a TM $M'$ with no move option such that $L(M) = L(M')$. This way is trivial, simply do not use $S$ move in the extended version.
In the other direction:
Given TM $M'$ with no move option, construct equivalent TM $M$ such that $L(M') = L(M)$.
Construct new transition function $\delta'$ for $M$ such that for each state with move $\in \{L, R\}$ in $\delta$ there is equivalent state in $\delta'$.
For each $S$ move in $\delta$ put two states in $\delta'$.
So for $\delta(q_x, a, b, S, q_y)$ put two states to $\delta'$, $\delta(q_x, a, b, L, q_n)$ and $\delta(q_n, *, *, R, q_y)$. The $q_n$ matches the acceptance with $q_x$.  If the $S$ move was accepting state, make the second state accepting. This gives the equivalence in the accepting states and the head (heads) position.
$L(M') = L(M)$, so these classes are equivalent.