Based on your question I assume the precision is very important in your lab. If so, then yes, your colleague is right. At least to some degree.
There are main three possibilities why this may happen. From your description the reason is purely physical, so option 1 on my list and I will explain it in more details. The other two are added just for completeness of the answer.
Why adding weights doesn't produce the sum of weights?
Consider these options.
1. If the substances mix (i.e. a result is a solution of the two substances) the volume of resultant solution is smaller than volumes of ingredients. The masses remain the same though. I'll explain this phenomenon in a bit. The reason is purely physical.
2. If the substances react and some of the resulting substances leaves the solution (usually it will be a gas that evaporates) the mass will reduce by the mass of removed substance. Usually the volume will also be smaller since on one hand some of the substance is removed and the resultant solution with whatever remains should follow the same principle as in 1. But the newly formed particles could in theory prevent that and cause the volume to actually increase. I can't think of any examples (or even if there exists some). The reason is mixed chemical and physical.
3. 979
If the substances react and as a result you get a soluble substance you end up with a solution of a new substance. In most (if not all) cases you have less particles to pack so your volume will drop with the mass remaining the same.If additionally the solvents mix you end up with added effect as in one so the difference can be even higher. This is a mix of chemical and physical reasons.
So what happens?
Your colleague refers specifically to point 1 on my list so I'll focus on it only.
In general the main reason is that the particles in solutions might have different size (and sometimes to a degree also shape, but usually it's less important factor) and as a result they can "pack" more densely so in the same volume you can fit more particles and as a result - more mass. The most extreme case is that the substances that are solvents themselves mix well with a significant volume reduction and the solved substance have just a tiny impact on the overall density.
As a result you end up with a mass being sum of masses but volume noticeably smaller than sum of volumes. And here enters the buoyancy on stage. The difference in volume causes a difference in weight. It's not big but if you need a huuuge precision it might make you fail meeting it.
So how big an impact can be?
In general the bigger the difference in particles size the stronger effect you'll notice.
I know nothing about the substances you're referring to but let's use something well known to most people. And something you can easily test yourself (and then consume the by-products to cool down your head a bit after thinking it all over ;-) ). There are two popular good solvents that mix rather easily - water and ethanol. I'll make a perfect case scenario where we mix pure water with pure ethanol. If you want to make an experiment you'll actually use some kind of solutions where those are solvents but the results will be noticeable for sure.
First let's calculate how much volume are we going to loose. Based on this water-ethanol mixtures density we'll calculate the difference for 40% alcohol solution (popularly known as vodka).
Pure water has a density of $0.998202 \text{ g}/\text{cm}^3$
Pure ethanol has a density of $0.79074 \text{ g}/\text{cm}^3$
40% ethanol in water solution (calculated by weight) has a density of $0.93684 \text{ g}/\text{cm}^3$
Those densities are based normalised atmosphere so weight based.
So let's take (by weight) $60 \text{ g}$ water and $40 \text{ g}$ ethanol. The volumes are respectively $60/0.998202 = 60.108074\text{ cm}^3$ of water and $40/0.79074 = 50.585527\text{ cm}^3$ ethanol.
A common sense would suggest we are going to have $100 \text{ g}$ of solution with a volume of $60.108074 + 50.585527 = 110.693601\text{ cm}^3$. Then we realise the density is going to be different, so $100 \text{ g}$ of solution that measures $100/0.93684 = 106.741813\text{ cm}^3$. But neither of those is correct.
To have accurate results we need to go through masses. 815