$p$-adic logarithm is a homomorphism, formal power series proof

It’s always possible that I have misunderstood the thrust of your question, but perhaps this argument will satisfy the preconditions you have set:

Set $G(x,y)=\log\bigl[(1+x)(1+y)\bigr]$ and $H(x,y)=\log(1+x)+\log(1+y)$. Take the derivative of each with respect to $x$. From $G$, you get $$ \frac1{(1+x)(1+y)}\frac\partial{\partial x}\bigl[(1+x)(1+y)\bigr]=\frac1{1+x}\,, $$ while from $H$ you get, of course, $\frac1{1+x}$. So $G$ and $H$ differ by a $y$-series: $$ \log\bigl[(1+x)(1+y)\bigr]=K(y)+\log(1+x)+\log(1+y)\,. $$ Now substitute $x=0$ and get $K=0$.