why do these two Meijer G functions not cancel each other?

Actually, the contours $C$ in the two integrals are different.

By definition, $C$ goes from $+infty$ through an clockwise path return to $+infty$, encircling all poles of $prod_{j=1}^mGamma(b_j-s)$ (each pole exactly once) and none pole of $prod_{j=1}^nGamma(1-a_j-s)$. In this case, the integral-contours of $textrm{G}^{4, 0}_{2, 4}Big({1,c+frac12atop c,c,c,1}Big|1Big)$ encircles poles:

$$c, c+1, c+2,dotsc, c+k,dotsc,$$

and poles:

$$0, -1, -2, dotsc, -k, dotsc,$$

must be outside the contours. Similar for $textrm{G}^{3, 1}_{2, 4}Big({1,c+frac12atop c,c,c,1}Big|1Big)$, we needs the contour encircles poles:

$$0, 1, 2, dotsc, k,dotsc,quad mathrm{and}quad c, c+1,c+2,dotsc,c+k,dotsc,$$

As you have found, $1,2,dotsc,n,dotsc$ are actually normal points of the integrand, so the contour can "deform" over these points. However, $0$ remains to be a pole, so the contour can not cross over it, which makes the expression obtain a non-zero value.

Thus, for any $cneq0,-1,-2,dotsc,-k,dotsc$ (by definition, they're not allowed), the sum of this two Meijer functions is:

$$-operatorname{Res}frac{[Gamma(c-s)]^3}{s,Gammabig(c+frac12-sbig)}Bigg|_{s=0} =-dfrac{[Gamma(c)]^3}{Gammabig(frac12+cbig)}.$$

Using Mathematica, we can check our answer:

Table[-MeijerG[{{},{1,c+1/2}},{{0,c,c,c},{}},1,-1]
    -MeijerG[{{1},{c+1/2}},{{c,c,c},{0}},1,-1]
    +Gamma[c]^3/Gamma[1/2+c]//N, 
    {c, SetPrecision[RandomReal[{1,2}, 10]
        +I RandomReal[{1, 2}, 10], 10]}
]//Column

Notice that in Mathematica

-MeijerG[{{__}, {__}}, {{__}, {__}}, _, -1]

is corresponding to the definition in Wikipedia(Bateman & Erdelyi).

Edit You may notice that when $c=-k-frac12$, $s=0$ will also be a normal point. Fortunately, the residue here as we have shown will be zero as we expected.