Maximality of the kernel of a quasifield

Question

Setting

A (left) quasifield is an algebraic structure $(Q,+,\cdot)$ such that

1. $(Q,+)$ is an abelian group. (As usual, we denote its identity element by $0$.)
2. Each equation $x\cdot a = b$ with $a,b\in Q$ and $a\neq 0$ has a unique solution $x\in Q$.
3. Each equation $a\cdot x = b\cdot x + c$ with $a,b,c\in Q$ and $a\neq b$ has a unique solution $x\in Q$.
4. Left distributive law: $a\cdot (b+c) = a\cdot b + a\cdot c$ for all $a,b,c\in Q$.

Intuitively, a left quasifield is a skew field minus multiplicative associativity minus the right distributive law.

Axiom 2 can be replaced by requiring that $(Q\setminus\{0\},\cdot)$ is a loop.

The kernel $K$ of a quasifield is defined as $$K = \{q \in Q \mid (ab)q = a(bq) \text{ and }(a+b)q = aq + bq\}.$$ It can be interpreted as the set of all elements of $Q$ which fulfill the missing associative and distributive law from the right side.

Now there is the following structure theorem:

Theorem

Let $Q$ be a left quasifield. Its kernel $K$ is a skew field and $Q$ is a right vector space over $K$.

My question

Is the kernel $K$ maximal with the above property?

More precisely: Is $K$ the maximal sub skew field of $Q$ such that $Q$ is a right vector space over $K$?

My thoughts

I don't know if the answer is "yes" or "no", but somehow I hope it is "yes".

So let $S$ be a sub skew field of $Q$ such that $Q$ is a right vector space over $S$. We would like to show $S \subseteq K$. By the definition of the kernel, for an arbitrary $s\in S$, we need to show $(a+b)s = as + bs$ and $(ab)s = a(bs)$ for all $a,b\in Q$.

The vector space axioms give us the first property, but unfortunately only a weaker variant of the second one, namely $(as')s = a(s's)$ for all $s,s'\in S$ and $a\in Q$. Let's call this the weak associative law.

My idea to show the more restrictive equality $(ab)s = a(bs)$ was to start with the identity $$(ab)1 = a(b1)$$ and transform it by using inverses in $S$. So assume $s\neq 0$ (otherwise, it's trivial). Let $t\in S$, $t\neq 0$ be the inverse of $s$. Plugging $1 = st$ into the above identity, we get $$(ab)(st) = a(b(st))$$ The weak associative law allows the transformation into $$((ab)s)t = a((bs)t)$$ Now I'm stuck.

If the right hand side could be transformed into $(a(bs))t$ (unfortunately that step is not covered by the weak associative law), then the claim would follow by right multiplication by $s$ and using the weak associative law and $ts = 1$.