The proposed inequality $\|\log(RS)\|\le\|\log(R)+\log)S)\|$ does not hold for all $R,S\in\mathrm{SO}(3)$ for which $\log(RS)$, $\log(R)$, and $\log(S)$ are defined. (In what follows, I'm taking the Frobenius norm to be $\|B\|^2 = \tfrac12\,\mathrm{tr}(B^TB)$, but this normalization clearly does not affect the argument.)
First, of all, notice that the function $B\mapsto \|\log(B)\|$ is actually a continuous function on $\mathrm{SO}(3)$, because of the easily checked identity $$ \cos\bigl(\|\log(B)\|\bigr) = \tfrac12\bigl(\mathrm{tr}(B)-1\bigr) $$ and the fact that $\cos^{-1}:[-1,1]\to[0,\pi]$ is a continuous, strictly decreasing function on $[-1,1]$. If the desired inequality were actually true, then we would have the inequality $$ \mathrm{tr}(RS) \ge 1+ 2\,\cos\bigl(\|\log(R)+\log(S)\|\bigr). $$ Now consider the two matrices $$ R = \exp\begin{pmatrix}0&-x&0\\x&0&0\\0&0&0\end{pmatrix} \quad\text{and}\quad S = \exp\begin{pmatrix}0&0&-y\\0&0&0\\y&0&0\end{pmatrix}, $$ where $0\le x,y < \pi$ (so that $R$, $S$, and $RS$ remain in the domain of the $\log$ function). Then, if the above inequality held, it would imply that $$ \cos x\cos y+\cos x + \cos y \ge 1 + 2\,\cos\left(\sqrt{x^2+y^2}\right). $$ However, this is not true. If you take $x=y$ in the above expression, then, indeed, the inequality holds for small $x\ge0$, but there is a value $x=y\approx 2.405181<\pi$ where both sides of this last inequality become equal and then the inequality reverses in the range $2.405182\le x=y <\pi$.
Thus, the original proposed inequality does not hold.