Let $A$ and $B$ be unital $C^*$-algebras. Let $u\in M_n(A)$ be a unitary representing an element $[u]\in K_1(A)$ and $p\in M_m(B)$ be a projection representing an element $[p]\in K_0(B)$. Then the unitary $$u\otimes p+1_A\otimes (1_B-p)$$ represets the product $[u]\times [p]\in K_1(A\otimes B)$ (see in Nigel Higson's and John Roes book "Analytic K-homology", proposition 4.8.3). This gives us a description of a product map $K_1(A)\times K_0(B)\to K_1(A\otimes B)$.

Now, let $A$ non-unital and $B$ unital, but $u$ and $p$ as above, where $u\in K_1(A)\cong K_1(A^+)$ with $A^+$ the unitization of $A$.

My question is: Is it possible to give an explicit description for $K_1(A)\times K_0(B)\to K_1(A\otimes B)$ as above for this case to map $[u]$ and $[p]$ to $K_1(A\otimes B)$ naturally?

I guess that this is not possible. My first try was to consider $$u\otimes p+1_{A^+}\otimes (1_B-p),$$ which represents an element in $K_1(A^+\otimes B)$. However, since $A\otimes B$ is non-unital, the product map should map to $K_1(A\otimes B)\cong K_1((A\otimes B)^+)$, but $ K_1((A\otimes B)^+)$ is not isomorphic to $K_1(A^+\otimes B)$. For $A$ nonunital and $B$ unital and nuclear, there is the following relation: there is a commutative diagram

$$\require{AMScd}\def\Z{\mathbb{Z}} \begin{CD} 0 @>>> K_1(A)\otimes K_0(B) @>>> K_1(A^+)\otimes K_0(B) @>>>> K_1(\mathbb{C})\otimes K_0(B) @>>> 0 \\ @. @VV V @VV\alpha V @VV\alpha V @. \\ 0 @>>>K_1(A\otimes B) @>>> K_1(A^+\otimes B) @>>> K_1(\mathbb{C}\otimes B) @>>> 0 \\ \end{CD} $$ where $K_1(\mathbb{C})=0$, $K_1(\mathbb{C}\otimes B)\cong K_1(B)$ and $K_1(A)\otimes K_0(B) \to K_1(A^+)\otimes K_0(B)$ is an isomorphism. If $B$ is nonnuclear, then the sequence in this diagram is not exact at $K_1(A\otimes B)$.