Yes, it has to do something with metallicity.
Brown dwarf SDSS J0104+1535 $\to 0.086\rm\, M_\odot\to[Fe/H]=-2.4$
Red dwarf EBLM J0555-57Ab $\to 0.081\rm\, M_\odot\to [Fe/H]=-0.24$
There is an article which answers just what you have asked. Let me quote the important part (I striked the wrong statement, helium is non-metal as well in astronomical sense):
Theory predicts that the mass cut-off for what constitutes a star is different for objects of different metallicity, which refers to the proportion of elements heavier than hydrogen the object contains.
For objects with a metallicity similar to that of the Sun, theory suggests that anything with less than 7.5% the mass of the Sun – or about 75 Jupiters – will be a brown dwarf.
In the case of NGC 6397, which has a metallicity 100 times lower than that of the Sun, the dividing line is expected to be at 8.3% the mass of the Sun, or about 83 Jupiters.
If we take a look at the red dwarf, it has approximately $10^{\text{[Fe/H]}}=10^{-0.24}=0.58$ times the metallicity of the Sun, so its metallicity is pretty Sun-like. Therefore, according to the article, anything above $0.075\rm\, M_\odot$ will be a star. Our red dwarf therefore lies in the star region.
However, if we take a look at the brown dwarf, in another article, it is stated that the calculated mass limit for a star with such metallicity is $0.088\rm\, M_\odot$, which is just slightly more than $0.086\rm\, M_\odot$ which is the real mass of the brown dwarf, so it still lies in the brown dwarf region.
But how does the metallicity influence the minimum mass for a star? ProfRob wraps this neatly in his answer. In short, the minimum mass is approximately given by
$$ M_{\rm min} \simeq 0.08 \left( \frac{\mu}{0.5} \right)^{-3/2} \left(\frac{\mu_e}{1.2}\right)^{-1/2}$$
where $\mu$ is the number of mass units per particle in core and $\mu_e$ is the number of mass units per electron in core which depends on the metallicity.