Given $u$, for every prime $p$ let $S_p = \{m \in (n,n+b)\colon p\mid m,\, p > m^u \}$. Then a lower bound on the quantity you want is $(b-1) - \sum_p \#S_p$. (Indeed, this is the exact value when $u<2$.) The smallest $p$ you have to consider is $n^{1/u}$, and an upper bound for $\#S_p$ is $\lceil\frac{b-1}p\rceil \le \frac{b-2}p+1$. You're therefore led to need to estimate $\sum_{n^{1/u} < p < n+b} 1$ and $\sum_{n^{1/u} < p < n+b} 1/p$. For both of these, the complete sums (where the lower bound is $1$) should appear explicitly in the work of Rosser–Schoenfeld, so just subtract two of those from each other to get the interval versions above.

Given $u$, for every prime $p$ let $S_p = \{m \in (n,n+b)\colon p\mid m,\, p^u > m \}$. Then a lower bound on the quantity you want is $(b-1) - \sum_p \#S_p$. (Indeed, this is the exact value when $u<2$.) The smallest $p$ you have to consider is $n^{1/u}$, and an upper bound for $\#S_p$ is $\lceil\frac{b-1}p\rceil \le \frac{b-2}p+1$. You're therefore led to need to estimate $\sum_{n^{1/u} < p < n+b} 1$ and $\sum_{n^{1/u} < p < n+b} 1/p$. For both of these, the complete sums (where the lower bound is $1$) should appear explicitly in the work of Rosser–Schoenfeld, so just subtract two of those from each other to get the interval versions above.