Well first when we restrict to the case when $A, B$ are filtered (see Bourbaki for example), in this case $\log_{*}$ always converges at $Id$ (as $Id=e+I_+$, $e$ being the unit for the convolution, it suffices to remark that $I_+^{*N}(h)=0$ for $N=N(h)$ large enough).

Now, in the general case, you can adapt the following computation to the algebra $H(A\otimes B)$ generated by the primitive elements of $A\otimes B$ where the series of $\log_{*_{12}}$ always converges.

For clarity, I note $A=A_1,B=A_2$ and and $e_i=1_{A_i}\circ \epsilon_i$.

Then
$$
\log_{*_{12}}(I_1\otimes I_2)=\log_{*_{12}}((I_1\otimes e_2)*_{12}(e_1\otimes I_2))=
$$
$$
\log_{*_{12}}(I_1\otimes e_2)+\log_{*_{12}}(e_1\otimes I_2)
$$

as the two terms $(I_1\otimes e_2), (e_1\otimes I_2)$ commute. Now
$\log_{*_{12}}(I_1\otimes e_2)=\log_{*_{1}}(I_1)\otimes e_2$ and
$\log_{*_{12}}(e_1\otimes I_2)=e_1\otimes\log_{*_{2}}(I_2)$.

Which, in view of $\left(\mathrm{Prim}A_1\right) \otimes k + k \otimes \left(\mathrm{Prim} A_2\right) \subseteq \mathrm{Prim}\left(A_1\otimes A_2\right)$, proves that $Prim(A_1\otimes A_2)=Prim(A_1)\otimes k+k\otimes Prim(A_2)$.

Which does your job.

**Addition** : *To answer your first question*, $I_1$ and $e_2$ are morphisms of bialgebras so $I_1\otimes e_2$ maps $Prim(A_1\otimes A_2)$ into $Prim(A_1\otimes A_2)$ and then $H$ into $H$ (in fact the image of $H(A_1\otimes A_2)$ is a subbialgebra of $H(A_1)\otimes k.1_{A_2}$).

*To answer the second point.* For a bialgebra let us denote $I^+=Id-e$ (the complement projector of $e$) and $H(?)$ the subalgebra generated by the primitive elements. One has, with the morphism of bialgebras
$$
(I_1\otimes e_2) : H(A_1\otimes A_2) \rightarrow H(A_1)\otimes k.1_{A_2}
$$
the intertwining
$$
(I_1^+\otimes e_2)\circ (I_1\otimes e_2)=(I_1\otimes e_2)\circ (I_1\otimes I_2)^+
$$
so, using series, we get
$$
(\log_{*_1}(I_1)\otimes e_2)\circ (I_1\otimes e_2)=(I_1\otimes e_2)\circ \log_{*_{12}}(I_1\otimes I_2)
$$
This is because, as a general principle, the intertwining intertwines the convolution. Let,
$$
\begin{matrix}
A & \stackrel{\varphi}{\longrightarrow} & B \cr
\downarrow && \downarrow \cr
A & \stackrel{\varphi}{\longrightarrow} & B
\end{matrix}
$$
with $\varphi$ a morphism of bialgebras and the down arrows $f,g$ such that $g\varphi=\varphi f$. Then, if the bialgebras are generated by primitive elements, if $f(1_A)=0,g(1_B)=0$ and if $S\in k[[x]]$ is a series, we have $S(g)\varphi=\varphi S(f)$. This is not difficult and argued in details (in particular the notion of summability and substitution) in my paper.

In conclusion, I think that
$$
Prim(A_1\otimes A_2)=Prim(A_1)\otimes k.1_{A_2}+k.1_{A_1}\otimes Prim(A_2)
$$
is true in full generality. One even does not have to suppose that $k$ is a field, only $\mathbb{Q}\subseteq k$ seems to be needed.

Do not hesitate to question and comment if something is unclear or wrong.

Regards