If $S$ is the surface illustrated below, do the Dehn twists about the red curves generate the mapping class group $\operatorname{MCG}(S,\partial S)$?

$\begingroup$ Hello, the picture has expired. $\endgroup$ – user135520 Oct 1 '17 at 3:33
No. If they did, then they would still generate the mapping class group of the closed surface that results from gluing a disc to the boundary component. However, in that surface they all commute with the hyperelliptic involution, which is not central for $g$ at least $3$.
In fact, Humphries proved that you need at least $2g+1$ Dehn twists to generate the mapping class group. This is contained somewhere in FarbMargalit's primer on mapping class groups.
In the closed surface, these Dehn twists actually generate the centralizer of the hyperelliptic involution, which is known as the hyperelliptic mapping class group. This is a theorem of Birman and Hilden, which can also be found somewhere in FarbMargalit.
Humphries arguments are in
Humphries, Stephen P. Generators for the mapping class group. Topology of lowdimensional manifolds (Proc. Second Sussex Conf., Chelwood Gate, 1977), pp. 44–47, Lecture Notes in Math., 722, Springer, Berlin, 1979.
The key argument for the $\ne 2g$ case is that the symplectic group $Sp(2g, \mathbf F_2)$ is not generated by $2g$ transvections (which maybe was well known). Steve and I joined up to present this argument and more in our paper
``Orbits under symplectic transvections II: the case $K=\bf F_2$'', Proc. London Math. Soc. (3) 52 (1986) 532556.
and Steve followed this up with much more work in this area.