Let $s, s_n\in\mathbb{R}$ and $\hat{s}_n$ be a random variable.
I have two concentration inequalities:
$$\mathbb{P}(\vert \hat{s}_n-s\vert > x)\leq a(n,x)$$ for all $n\geq1$ and $x>0$; and $$\mathbb{P}(\vert \hat{s}_n-s_n\vert > x)\leq b(n,x)$$ for all $n\geq1$ and $x>0$.
Is there a way to bound $\vert s_n - s\vert$?
Yes, there is a way. Let $$X:=\hat s_n-s,\quad h:=s_n-s,\quad a(x):=a(n,x),\quad b(x):=\limsup_{y\uparrow x}b(n,y).$$ We have $$P(|X|>x)\le a(x)\quad\text{and}\quad P(|X-h|\ge x)\le b(x)$$ for all $x>0$, and we need to bound $|h|$ in terms of the functions $a$ and $b$.
The simple but key observation is 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 $$c^{-1}(u):=\sup\{x\ge0\colon c(x)\ge u\}$$ 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 $\sigma>0$, then the upper bound $h_*$ on $|s_n-s|$ is $2\sigma\sqrt{\ln4}$, which is proportional to $\sigma$.
