Use $\mathbb{P}(\vert \hat{s}_n-s\vert > x)\leq a(n,x)$ and $\mathbb{P}(\vert \hat{s}_n-s_n\vert > x)\leq b(n,x)$ to bound $\vert s_n - s\vert$ 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 $\vert s_n - s\vert$?

Assuming that $\hat{s}_n$ is integrable,
\begin{align}
|s_n-s|&\le \mathsf{E}|s_n-\hat{s}_n|+\mathsf{E}|s-\hat{s}_n| \\
&=\int_0^{\infty}\mathsf{P}(|s_n-\hat{s}_n|>x)\,dx+\int_0^{\infty}\mathsf{P}(|s-\hat{s}_n|>x)\,dx \\
&\le \int_0^{\infty}[a(n,x)+b(n,x)]\,dx.
\end{align}
