This is a very common theme in enumerative combinatorics. You can find a lot of examples with the Google search "no bijective proof" (with quotes).

First, I can say something about why you might care about bijective proofs. Combinatorial species are certainly a nice theory, but they are a fairly specific and elaborate answer related to generating functions. A more general reason is that a bijective proof categorifies an equality in combinatorics to the category of sets. In other words, it promotes an equality $|A| = |B|$ to an isomorphism $A \cong B$. In my opinion, it is just as important to find any other categorification, for instance to the category of vector spaces. Instead of showing that two sets are the same size with a bijection, you would show that they are the same size using an invertible matrix.

Second, of the many examples, I can name one that I encountered. This example is interesting because the objects in question seem very similar. Recall that an alternating sign matrix is a matrix whose non-zero entries in each row and column alternate between $1$ and $-1$, and such that the first and last non-zero entry in each row and column is $1$. One interesting subclass is the ASMs of order $2n+1$ which are symmetric about a vertical line. Another interesting subclass is the ASMs of order $2n$ which are diagonally symmetric and have 0s on the diagonal. (ASMs of either type of the opposite parity do not exist.) The first class was discovered by David Robbins and I found the second class. I proved David's product formula for the first class and I established the same product formula for the second class. So these two classes of ASMs are equinumerous, but no bijective proof is known.

Here is another interesting example in the same vein. A cyclically symmetric, self-complementary plane partition (CSSCPP) is equivalent to a tiling of a regular hexagon of order $2n$ by unit lozenges, which is invariant under 60 degree rotation. Here a unit lozenge is two unit equilateral triangles stuck together. A totally symmetric, self-complementary plane partition (TSSCPP) is the same thing except with full dihedral symmetry. (I make the size even because otherwise there aren't any plane partitions with the imposed symmetry.) The formulas for both classes were also conjectured by David Robbins; George Andrews proved his conjecture for TSSCPPs and I proved the conjecture for CSSCPPs. In particular, the number CSSCPPs of a fixed size is the square of the number of TSSCPPs, but no one knows a good bijection.

The single most striking thing that David Robbins found was that the number of TSSCPPs, which are plane partitions with full symmetry, equals the number of ASMs with no imposed symmetry. No bijective proof of that is known either. On the positive side, Doron Zeilberger's proof of the ASM conjecture, and his later paper on refined ASMs, could be steps towards one because they equate certain generalizations and refined enumerations. However, alternating-sign matrices look totally different from plane partitions. In my opinion, the most frustrating case is when we can't even match like to like.