Archimedes' principle tells us that the upwards force on an object immersed in a fluid is equal to the weight of fluid displaced.
So let's take your initial experiment. You don't tell us the volume of your object, but you do give its weight, $W_g = 100$ g, and density, $\rho_g = 2.6$ g/cm$^3$, so the volume is:
$$ V = \frac{W_g}{\rho_g} = 38.46 cm^3 $$
When you put this in water the volume of water displaced is the same as the above volume, so the mass of water displaced is:
$$ M_w = V\rho_w = \frac{W_g}{\rho_g} \rho_w = 38.46g $$
and hence the effective mass of your glass object is 100 - 38.46 = 61.54g as you say.
If the density of water changes to 1.010 g/cm$^{3}$ just use the equation above but set $\rho_w = 1.010$, and you will indeed get the effective weight of the glass equal to 61.15g. Though if you want to be really accurate you should note that changing the temperature will change the density of the glass as well, so $\rho_g$ would be slightly different as well.
From a unit analysis point of view mass and weight are different kinds of things: mass is a quantity of material substance and weight is a force. Things of a different kind can never be the same.
That said, historically the distinction was not always recognized and traditional systems of weight-and-measure (including the original metric system, but not SI) use the same unit for both.
In those systems of measurement, there have the same numeric value on Earth by construction.
Best Answer
As you rightly pointed out, the fact that 67P is oddly-shaped should alter its gravitational attraction on various parts of the comet.
That said, if we were to go by Wikipedia in a rather off-hand manner, we find that the lander is $100 kg$ (as @fibonatic rightly pointed out) and 67P has an acceleration due to gravity of $\textbf{g'} = 10^{-3} m/s^2$. Its weight $W$ would therefore be simply a calculation of $W = m\,\textbf{g'}$, giving us $W = \frac{10^2}{10^3} kg = 0.1 kg$ or $100 \,\,\verb+earth+\, g$.
[P.S. I will update this answer with better sources than Wikipedia as soon as I find time.]Edit 1: This ESA webpage seems to confirm the figures.
Edit 2: Calculating $\textbf{g'}$
I made some calculations: Using the formula $\textbf{F} = M\,\textbf{g'} = \frac{GmM}{r^2} \Rightarrow \textbf{g'} = \frac{GM}{r^2}$ we can calculate the acceleration due to gravity on 67P (m being the comet's mass and M that of our lander). The above ESA page gives us this figure:
Seeing how the dimensions vary wildly, I decided to consider a mean of, say, 3.5km as diameter and 1.75km as $r$. 67P's mass is, of course, $10^{13}\,kg$ which gives us, $$ \textbf{g'} = \frac{6.67 \times 10^{-11} \times 10^{13}}{\left( 1.75 \times 10^3 \right)^2} \approx 10^{-3} ms^{-2} $$ A more precise answer, is, of course, $0.217 \times 10^{-3} ms^{-2}$ but since we have been very liberal in our assumptions of mass and radius, I think we ought to simply consider the order of magnitude, $10^{-3} ms^{-2}$. This pdf file contains some simulation data that agrees with our result.