$\ln(a+b)\leq\ln(a)+\ln(b)$

You are correct. This indeed fails if $a=1$ and $b\in\mathbb{N}$, as $\log(1)=0$, so the inequality becomes $$\log(1+b)\leq\log(b)$$ which is clearly false. However, if we strengthen the premise to $a,b>1$, then this inequality holds.

Your strengthening to $1<b<a$ omits the equality case $a=b=2$, as $$\log(2+2) = \log(2\cdot 2) = \log(2)+\log(2)$$