First, exclude all the terms such that $a_i=b_i$, since these obviously do not affect the inequality. We show that the sequences can be made so that $(b_i)$ majorises $(a_i)$, and then we apply Karamata's inequality.
Let $a_1,b_1,\cdots a_p,b_p\geq 0$ and the remaining terms negative, and the same for $b_i.$ Since $|a_i|<|b_i|:$
$$a_1< b_1$$
$$a_1+a_2< b_1+b_2$$
$$\cdots$$
$$a_1+\cdots +a_p< b_1+\cdots +b_p$$
There are two possibilities here. Either $n=p+1,$ in which case adding $a_{p+1},b_{p+1}$ yields equality and we are done. Or $n>p+1$, in which case we must still have:
$$a_1+\cdots+a_{p+1}<b_1+\cdots +b_{p+1}$$
This follows from a simple contradiction: suppose $a_1+\cdots+a_{p+1}\geq b_1+\cdots +b_{p+1}$ then $a_1+\cdots +a_{n}>b_1+\cdots+b_n$ since $a_j>b_j$ for $j>p$. This is false, hence inductively:
$$a_1+\cdots +a_i\leq b_1+\cdots +b_i,\;\;1\leq i\leq n$$
Thus $(b_i)$ majorises $(a_i)$, and for any convex $f:$
$$f(a_1)+f(a_2)+\cdots +f(a_n)\leq f(b_1)+f(b_2)+\cdots +f(b_n)$$
This problem is simply the case $f=\exp$.