The simplest explanation is that it is a mistake. One can however complete the proof as follows:

If $X$ is large enough then $u(m) \in \{ 0,1,2 \}$ for all $m$ since $e < 3$. Thus the number of distinct primes in $L$ dividing $m$ is at least $u(m)$, unless $u(m) = 2$ and $m$ has only one prime factor in $L$, in which case $m = p^2 q$. 

Let us count pairs of integers $(p,q)$ where $p > R := c_{8} X^{\frac{1}{e}}$ and $m < p^2 q \leq m+t$. This it at most
$$
\sum_{q \leq X R^{-2}} \mathrm{number \ of \ squares \ in \ } ]mq^{-1},(m+t)q^{-1} ] \\
\leq \sum_{q \leq X R^{-2}} \left( \frac{t}{\sqrt{qm}} + O(1) \right) \\
\ll \frac{t}{\sqrt{R}} \sqrt{X R^{-2}} + X R^{-2} \\
\ll t X^{-\alpha} +X^{\frac{1}{e} - 2 \alpha}
$$
where $\alpha = \frac{3-e}{2e} >0$.

Thus the number of distinct prime factors of $\prod_{j=1}^{t} (m+j)$ in $L$ is at least $t - O(t X^{-\alpha} +X^{\frac{1}{e} - 2 \alpha})$. By choosing $t$ of size $X^{\frac{1}{e}}$, one thus gets $\geq t - O(t X^{-\alpha})$ distinct prime factors $>t$. By adding the $\pi(t)$ primes $\leq t$, we get $\geq t$ distinct prime factors (for $X$ large enough).