One short way of getting the desired isomorphism is
\begin{align*}
(K \otimes A) \otimes_K (K \otimes B)
&\cong
(A \otimes K) \otimes_K (K \otimes B)
\\
&\cong
A \otimes (K \otimes_K (K \otimes B))
\tag{$\ast$}
\\
&\cong
A \otimes K \otimes B
\\
&\cong
K \otimes A \otimes B
\\
&\cong
K \otimes (A \otimes B)
\end{align*}
where we use for $(\ast)$ the (generalized) associativity over the tensor product.
To explicitely construct the mutually inverse homomorphisms
\begin{align*}
F
&\colon
(K \otimes A) \otimes_K (K \otimes B)
\to
K \otimes_K (A \otimes B) \,,
\\
G
&\colon
K \otimes_K (A \otimes B)
\to
(K \otimes A) \otimes_K (K \otimes B)
\end{align*}
we can proceed like you suspect.
To construct the bilinear map
\begin{align*}
f
\colon
(K \otimes A) \times (K \otimes B)
&\to
K \otimes (A \otimes B) \,,
\\
(\lambda \otimes a, \mu \otimes b)
&\mapsto
(\lambda \mu) \otimes (a \otimes b)
\end{align*}
we start with the $\mathbb{Z}$-multilinear map
\begin{align*}
f''
\colon
K \times A \times K \times B
&\to
K \otimes (A \otimes B) \,,
\\
(\lambda, a, \mu, b)
&\mapsto
(\lambda \mu) \otimes (a \otimes b) \,.
\end{align*}
For fixed $(\lambda, a) \in K \times A$ we get a $\mathbb{Z}$-bilinear map
\begin{align*}
f''_{(\lambda,a)}
\colon
K \times B
&\to
K \otimes (A \otimes B) \,,
\\
(\mu, b)
&\mapsto
(\lambda \mu) \otimes (a \otimes b)
\end{align*}
and thus an induced $\mathbb{Z}$-linear map
\begin{align*}
f'_{(\lambda,a)}
\colon
K \otimes B
&\to
K \otimes (A \otimes B) \,,
\\
\mu \otimes b
&\mapsto
(\lambda \mu) \otimes (a \otimes b) \,.
\end{align*}
The mapping $(\lambda,a) \mapsto f''_{(\lambda,a)}$ is itself $\mathbb{Z}$-bilinear by the $\mathbb{Z}$-multilinearity of $f''$, whence the mapping $(\lambda,a) \mapsto f'_{(\lambda,a)}$ is $\mathbb{Z}$-linear.
The resulting map
\begin{align*}
f'
\colon
K \times A \times (K \otimes B)
&\to
K \otimes (A \otimes B) \,,
\\
(\lambda, a, \mu \otimes b)
&\mapsto
f'_{(\lambda,a)}(\mu \otimes b)
=
(\lambda \mu) \otimes (a \otimes b) \,.
\end{align*}
is therefore $\mathbb{Z}$-trilinear.
For fixed $t \in K \otimes B$ we now get again a $\mathbb{Z}$-bilinear map
\begin{align*}
f'_{t}
\colon
K \times A
&\to
K \otimes (A \otimes B) \,,
\\
(\lambda, a)
&\mapsto
f'(\lambda, a, t) \,.
\end{align*}
that results again in a $\mathbb{Z}$-linear map
\begin{align*}
f_t
\colon
K \otimes A
&\to
K \otimes (A \otimes B) \,,
\\
\lambda \otimes a
&\mapsto
f'(\lambda, a, t) \,.
\end{align*}
It follows from the $\mathbb{Z}$-trilinearity of $f'$ that the mapping $t \mapsto f'_t$ is $\mathbb{Z}$-linear, whence the mapping $t \mapsto f_t$ is $\mathbb{Z}$-linear.
This means that the now resulting map
\begin{align*}
f
\colon
(K \otimes A) \times (K \otimes B)
&\to
K \otimes (A \otimes B) \,,
\\
(s, t)
&\mapsto
f_t(u)
\end{align*}
is $\mathbb{Z}$-bilinear.
For $s = \lambda \otimes a$ and $t = \mu \otimes b$ we have by construction
\begin{align*}
f(\lambda \otimes a, \mu \otimes b)
&=
f_{\mu \otimes b}(\lambda \otimes a)
\\
&=
f'(\lambda, a, \mu \otimes b)
\\
&=
f'_{(\lambda, a)}(\mu \otimes b)
\\
&=
f''_{(\lambda, a)}(\mu, b)
\\
&=
(\lambda \mu) \otimes (a \otimes b) \,.
\end{align*}
Now we need to check that the $\mathbb{Z}$-bilinear map $f$ is already $K$-bilinear.
For this we still need that
$$
f(\kappa \cdot s, t)
=
\kappa \cdot f(s,t)
=
f(s, \kappa \cdot t)
$$
for all $\kappa \in K$ and $s \in K \otimes A$, $t \in K \otimes B$.
We may assume that $s$,$t$ are simple tensors $s = \lambda \otimes a$ and $t = \mu \otimes b$ and then
\begin{align*}
f(\kappa \cdot (\lambda \otimes a), (\mu \otimes b))
&=
f((\kappa \lambda) \otimes a, \mu \otimes b)
\\
&=
(\kappa \lambda \mu) \otimes (a \otimes b)
\\
&=
\kappa \cdot ((\lambda \mu) \otimes (a \otimes b))
\\
&=
\kappa \cdot f(\lambda \otimes a, \mu \otimes b) \,.
\end{align*}
The other equation can be shown in the same way.
It follows that $f$ induces a $K$-linear map
\begin{align*}
F
\colon
(K \otimes A) \otimes_K (K \otimes B)
&\to
K \otimes (A \otimes B) \,,
\\
(\lambda \otimes a, \mu \otimes b)
&\mapsto
(\lambda \mu) \otimes (a \otimes b) \,.
\end{align*}
To construct the inverse $G$ of $F$ we can play the same game:
We start with a $\mathbb{Z}$-trilinear map
\begin{align*}
g''
\colon
K \times A \times B
&\to
(K \otimes A) \otimes_K (K \otimes B) \,,
\\
(\lambda, a, b)
&\mapsto
(\lambda \otimes a) \otimes (1 \otimes b) \,.
\end{align*}
(Note that it dosen’t matter in which tensor factor $K \otimes A$ or $K \otimes B$ we put $\lambda$.)
Then we get
\begin{align*}
g'_{\lambda}
&\colon
A \times B
\to
(K \otimes A) \otimes_K (K \otimes B) \,,
\\
g_{\lambda}
&\colon
A \otimes B
\to
(K \otimes A) \otimes_K (K \otimes B) \,,
\\
g
&\colon
K \times (A \otimes B)
\to
(K \otimes A) \otimes_K (K \otimes B) \,,
\\
G
&\colon
K \otimes (A \otimes B)
\to
(K \otimes A) \otimes_K (K \otimes B) \,,
\end{align*}
It can then be checked on simple tensors that the maps $F$ and $G$ are mutually inverse.
Another way to construct $F$ is by the tensor-$\operatorname{Hom}$-adjunction:
The map
$$
A \times B
\to
A \otimes B \,,
\quad
(a,b)
\mapsto
ab
$$
is $\mathbb{Z}$-bilinear and hence corresponds a $\mathbb{Z}$-linear map
$$
A
\to
\operatorname{Hom}_{\mathbb{Z}}(B, A \otimes B) \,.
$$
Every homomorphism $B \to A \otimes B$ can be extended to a $K$-linear map $K \otimes B \to K \otimes (A \otimes B)$, giving a $\mathbb{Z}$-linear map
\begin{align*}
\operatorname{Hom}_{\mathbb{Z}}(B, A \otimes B)
&\to
\operatorname{Hom}_K(K \otimes B, K \otimes (A \otimes B)) \,,
\\
h
&\mapsto
\mathrm{id}_K \otimes h \,.
\end{align*}
By composing the above two $\mathbb{Z}$-linear maps we get a $\mathbb{Z}$-linear map
$$
A
\to
\operatorname{Hom}_K(K \otimes B, K \otimes (A \otimes B)) \,.
$$
The right hand side is a $K$-module, so by extension of scalars we get a $K$-linear map
$$
K \otimes A
\to
\operatorname{Hom}_K(K \otimes B, K \otimes (A \otimes B)) \,.
$$
This $K$-linear map corresponds to a $K$-bilinear map
$$
(K \otimes A) \times (K \otimes B)
\to
K \otimes (A \otimes B) \,,
$$
which is precisely $f$, and hence to a $K$-linear map
$$
(K \otimes A) \otimes_K (K \otimes B)
\to
K \otimes (A \otimes B) \,,
$$
which is precisely $F$.
