Yes, there is a way. Let 
$$X:=\hat s_n-s,\quad h:=s_n-s,\quad a(x):=a(n,x),\quad b(x):=b(n,x).$$
We have 
$$P(|X|>x)\le a(x)\quad\text{and}\quad P(|X-h|>x)\le b(x)$$
for all $x>0$, and we need to bound $|h|$ in terms of the functions $a$ and $b$. 

Note that the event $\{|X-h|<|h|/2\}$ implies the event $\{|X|>|h|/2\}$. So,
$$a(|h|/2)\ge P(|X|>|h|/2)\ge P(|X-h|<|h|/2)\ge1-b(|h|/2),$$
whence $c(|h|/2)\ge1$, where $c:=a+b$. So, 
$$|s_n-s|=|h|\le h_*:=2c^{-1}(1),$$
where $c^{-1}$ is the generalized inverse of the function $c\colon[0,\infty)\to\mathbb R$ given by the formula 
$$c^{-1}(u):=\sup\{t\ge0\colon c(t)\ge u\}$$
for $u\in(0,1]$; 
if $c$ is continuously and strictly decreasing from $c(0)\ge1$ to $c(\infty-)=0$, then $c^{-1}$ is the usual inverse of the function $c$. 

If e.g. $a(x)=b(x)=2e^{-x^2/\sigma^2}$ for some real $s>0$, then the upper bound $h_*$ on $|s_n-s|$ is $2\sigma\sqrt{\ln4}$, which is proportional to $\sigma$.