If $A\times (-)$ is a left adjoint, it must commute with colimits, and in particular, $\oplus$.

That is, write $B_1\oplus B_2$ as $colim_i B_i$, then $A\times colim_iB_i\cong colim_i(A\times B_i)$, from which the result follows.

If $A\times (-)$ is a left adjoint, it must commute with colimits, and in particular, $\oplus$.

That is, write $B_1\oplus B_2$ as $colim_i B_i$, then $A\times colim_iB_i\cong colim_i(A\times B_i)$, from which the result follows.

Since $A\times (-)$ must preserve colimits, also notice that the initial object is the colimit over the empty diagram, so by preservation of colimits, we win again!