Is there anything known about when two links have the same Jones polynomial (beyond a calculated list of small actual examples)? The first thing I would try is to compute the (formal - you would have something*split-horizontal+something*split-vertical) Jones polynomial of (4-)tangles and look there for two tangles with same Jones polynomial. Any instance would then generate an infinite example family. The snag would be, of course, that I'd first need a tangle table - my own goes to meagre 6 crossings. Aaaand the computation can't be automated that good. Still - any takers?
Probably the best way to produce infinite families of links with same Jones polynomial is by Conway mutation, this operation does not alter the HOMFLY polynomial either. A good example of this is given by the family of pretzel links. Take a look at the answer of a question of mine here:
By using satellites of the Hopf link it is possible to produce an infinite family of links with the same Jones polynomial of the trivial link. This was done by Shalom Eliahou, Louis H. Kauﬀman and Morwen B. Thistlethwaite:
Liam Watson generalized a construction of Kanenobu to produce infinitely many pairs of knots with the same Jones polynomial (and Khovanov homology) but distinct HOMFLYPT polynomials, so they are not mutants. See the references below.
MR2287438 Watson, Liam. Any tangle extends to non-mutant knots with the same Jones polynomial. J. Knot Theory Ramifications 15 (2006), no. 9, 1153–1162.
MR2350287 Watson, Liam. Knots with identical Khovanov homology. Algebr. Geom. Topol. 7 (2007), 1389–1407.
It is well-known that the Jones polynomial of the connected sum of $L_1$ and $L_2$ is exactly the product of the Jones polynomial of $L_1$ and $L_2$. However for a pair of links $L_1$ and $L_2$, the connected sum is not well-defined. For example assume $L_1=K_1\cup K_2$ and $L_2=K_3\cup K_4$, then you can connect $K_1$ with $K_3$, or connect $K_2$ with $K_4$. In general they are different links, but the Jones polynomial of them are the same, both equals to $V(L_1)V(L_2)$.