The hazard function $\lambda(t)$ is related to the survival function $S(t)$ as
\begin{align}\lambda(t) &= \frac{-\frac{\mathrm d(St)}{\mathrm dt}}{S(t)}\\
S(t) &= \exp\left(-\int_0^t\lambda(s)\,\mathrm ds\right)\end{align}
and so
\begin{align}S_\alpha(T) &= S(T+\alpha)\\
&= \exp\left(-\int_0^{T+\alpha}\lambda(s)\,\mathrm ds\right)
\\&= \exp\left(-\int_0^{T}\lambda(s)\,\mathrm ds
- \int_T^{T+\alpha}\lambda(s)\,\mathrm ds\right)\\
&= S(T)\cdot \exp\left(-\int_T^{T+\alpha}\lambda(s)\,\mathrm ds\right)\end{align}
where you calculated everything except the very last step in writing
your question. Unfortunately, no further simplification is possible
without more specific information about the hazard function. But, since
we do know that $\lambda(t) \geq 0$ for all $t$, we do get the cold
comfort of knowing that $S_\alpha(T)$ is a nonincreasing function
of $\alpha$.