The mapping $\iota \otimes S$ on the multiplier algebra Let $A$ be a non-degenerate algebra with multiplier algebra $M(A)$. Let $S: A \to M(A)$ be an antimultiplicative linear map, i.e.
$$S(ab) = S(b)S(a).$$
Consider the mapping
$$\iota \otimes S: A \otimes A \to A \otimes M(A) \subseteq M(A \otimes A).$$
Does it extend to a map $\iota \otimes S: M(A \otimes A) \to M(A \otimes A)$?
The results that I know for extending linear maps to the multiplier algebra require (anti)multiplicativity to make sure the extension is well-defined. Here however, $\iota \otimes S$ is neither multiplicative nor antimultiplicative, but I guess there should be some way around this problem.

To give some context, in Van Daele's paper "An algebraic framework on group duality", in proposition 3.11, he considers an expression $(\iota \otimes S)(\Delta(q))$ where $S: A \to A$ is the antipode on the regular multiplier Hopf algebra $(A, \Delta)$. Since $\Delta(q) \in M(A \otimes A)$, we need to be able to make sense of $\iota \otimes S$ on the multiplier algebra $M(A \otimes A)$.
Here is a screenshot:

 A: Firstly, in An Algebraic Framework for Group Duality van Daele only considers regular multiplier Hopf algebras, so $S(A)\subseteq A$, and $S$ is a bijection, see Proposition 2.7.
Then, given such $S$, we can extend it to a map $M(A)\rightarrow M(A)$ by defining, for example, $S(x) a = S( S^{-1}(a)x )$ and $a S(x) = S( x S^{-1}(a) )$, for $a\in A, x\in M(A)$.  See the discussion by van Daele on pages 333-334.
We consider Proposition 3.11 (for example, as it is in the OP). Here all elements are in $A$, and we have a formula like
$$ (x\otimes p)(\iota\otimes S)\Delta(q). $$
This, however, is equal to
$$ (1\otimes p)(\iota\otimes S)\big( (x\otimes 1)\Delta(q) \big), $$
and this makes perfect sense, as $(x\otimes 1)\Delta(q) \in A\otimes A$.  In fact, examining van Daele's proof, it's fairly clear that this must be the interpretation being used.
Notice that if we set $(x\otimes 1)\Delta(q) = \sum_i a_i \otimes b_i$ then
$$ (1\otimes p)(\iota\otimes S)\big( (x\otimes 1)\Delta(q) \big)
= \sum_i a_i \otimes pS(b_i) = \sum_i a_i \otimes S(b_i S^{-1}(p))
= (\iota\otimes S)\big( (x\otimes 1)\Delta(q)(1\otimes S^{-1}(p) \big)
= (x\otimes 1)(\iota\otimes S)\big( \Delta(q)(1\otimes S^{-1}(p) \big)
, $$
and so we find that
$$ (1\otimes p)(\iota\otimes S)\Delta(q)
= (\iota\otimes S)\big( \Delta(q)(1\otimes S^{-1}(p) \big). $$
I think it is correct to believe that $(\iota\otimes S)\Delta(q)$ does not make (complete) sense in isolation, but we can always argue in ways similar to the above.
