I'm not sure that your elementary argument is correct. Not only does the PNT not bite fast enough, but also the fact that every prime less than $k$ must divide $q$ gives only that $S(k)>(k-1)\prod_{p_i< k}p_i$, or (using the primorial notation) $S(k)>(k-1)(k-1)\#$. It is easy to show that $n\#>n$, and hence we get the lower bound
$$S(k)>(k-1)^2$$
Let $l=\lfloor \log_2(k-1)\rfloor$. Then it is slightly harder (e.g. see http://godplaysdice.blogspot.com/2007/09/hand-waving-asymptotics-of-primorial.html) to show that
$$(k-1)\#>\frac{(k-1)^l}{2^{l(l+1)/2}} $$
This gives us $$S(k)>\frac{(k-1)^{l+1}}{2^{l(l+1)/2}}>(k-1)^{(l+1)/2} $$ I'm not sure (barring using improved bounds for the primorial) whether these methods can get anything better.
You can regain your original $\prod_{p_i\leq k}p_i$ if you can rule out the possibility that the AP can start with $k$ itself, and this then gives you a lower bound $$S(k)>(k-1)k^{(l-1)/2}$$ where $l=\lfloor\log_2k\rfloor$.
Update: Of course, you can also gain a slight improvement by noting that the first term of the AP must be at least k, and hence
$$S(k)>k+(k-1)^{(l+1)/2}$$
since if we write our AP as $a,a+q,...,a+(k-1)q$then we're trying to find a lower bound for $a+(k-1)q$. This combines our best known bounds for both $a$ and $q$. I suspect that asymptotically $q$ will behave like $k\#$, and so the $e^k$ is best possible, and the $(k-1)$ factor is fixed, so any improvements must come from better lower bounds for $a$, but these would be negligible compared to $e^k$. In summary, I think your given lower bound is essentially best possible.
