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I don't know of a resource with such a list, though you could easily make one with programs such as GAP or Magma. Since it is reasonably easy and interesting to analyze this problem, I'll outline how I suggest thinking about the examples:

Let ⟨a⟩ act on the normal subgroup ⟨b⟩ in the semi-direct product ⟨a,b⟩. Then z = aibj is in the center of ⟨a,b⟩ if and only if it commutes with both a and b. Just writing it out you get a⋅(aibj) = ai+1bj and (aibj)⋅a = ai+1(bj)a, so bj has to be centralized by a. Similarly to commute with b, b has to be centralized by ai.

You could try to use this in two directions: make ⟨a⟩ centralize part of ⟨b⟩, or let the kernel of the action of ⟨a⟩ on ⟨b⟩ be non-trivial (and proper).

The first direction runs into some trouble, but does let you produce pretty obvious examples. As a concrete family, consider:

Let b have order 3k for k an integer coprime to 6, and let a of order 2 act on b as ba = b(k2+1). Then ⟨a,b⟩ has center ⟨bk⟩.

This example is not particularly satisfying since it is just the direct product S3×k. For the verification, check that (b3)a = b(3k2+3) = b3 since 3 ≡ 3k2+3 mod 3k, but (bk)a = bk(k2+1) = (bk)2 since 2 ≡ k2+1 mod 3. This might make a good example to show that "a" and "b" are not uniquely defined by the abstract isomorphism type of the group meeting your requirements (since the "k" could just as easily have been part of the "a" as part of the "b").

Unfortunately you cannot do much better than this using the first direction, since if ⟨a⟩ centralizes a subgroup of ⟨b⟩ of prime order p, then ⟨a⟩ centralizes the entire Sylow p-subgroup of ⟨b⟩, and you get a direct product. This follows from your coprime hypothesis.

The other direction looks a bit different. Take any example semi-direct product ⟨A,b⟩ that has ⟨b⟩ normal, |a| A| and |b| are coprime, and Z(⟨A,b⟩) a proper subgroup. Let ⟨a⟩ be a cyclic group of order k|A| where k is coprime to ⟨b⟩, and define ba = bA. Then the center of ⟨a,b⟩ is still a proper subgroup and contains ⟨a|A|⟩ of order k. More concretely:

Let b have odd order m and let a have order 2n+1 and let a act on b as ba = b−1. Then ⟨a,b⟩ has center ⟨a2⟩ of order 2n.

You can combine the two directions, but again because the first direction trivializes, the combined version looks pretty silly. Just take the direct product any example of the second direction and a cyclic group of order coprime to its |a|. This exhausts all possibilities, but I suspect most people would be irritated with the direct product aspect and only consider the second direction valid.

If you relax the coprime hypothesis, then you get another type of example that is reasonably interesting: take the Sylow p-subgroup of the holomorph of a cyclic p-group. The center will have order exactly p. If ⟨b⟩ has order pn+1, then ⟨a⟩ will have order pn, ⟨a,b⟩ will have order p2n+1, and the center of ⟨a,b⟩ will be ⟨bpn⟩ of order p. The action of a on b is easy to write down: ba = b1+p.

1

I don't know of a resource with such a list, though you could easily make one with programs such as GAP or Magma. Since it is reasonably easy and interesting to analyze this problem, I'll outline how I suggest thinking about the examples:

Let ⟨a⟩ act on the normal subgroup ⟨b⟩ in the semi-direct product ⟨a,b⟩. Then z = aibj is in the center of ⟨a,b⟩ if and only if it commutes with both a and b. Just writing it out you get a⋅(aibj) = ai+1bj and (aibj)⋅a = ai+1(bj)a, so bj has to be centralized by a. Similarly to commute with b, b has to be centralized by ai.

You could try to use this in two directions: make ⟨a⟩ centralize part of ⟨b⟩, or let the kernel of the action of ⟨a⟩ on ⟨b⟩ be non-trivial (and proper).

The first direction runs into some trouble, but does let you produce pretty obvious examples. As a concrete family, consider:

Let b have order 3k for k an integer coprime to 6, and let a of order 2 act on b as ba = b(k2+1). Then ⟨a,b⟩ has center ⟨bk⟩.

This example is not particularly satisfying since it is just the direct product S3×k. For the verification, check that (b3)a = b(3k2+3) = b3 since 3 ≡ 3k2+3 mod 3k, but (bk)a = bk(k2+1) = (bk)2 since 2 ≡ k2+1 mod 3. This might make a good example to show that "a" and "b" are not uniquely defined by the abstract isomorphism type of the group meeting your requirements (since the "k" could just as easily have been part of the "a" as part of the "b").

Unfortunately you cannot do much better than this using the first direction, since if ⟨a⟩ centralizes a subgroup of ⟨b⟩ of prime order p, then ⟨a⟩ centralizes the entire Sylow p-subgroup of ⟨b⟩, and you get a direct product. This follows from your coprime hypothesis.

The other direction looks a bit different. Take any example semi-direct product ⟨A,b⟩ that has ⟨b⟩ normal, |a| and |b| are coprime, and Z(⟨A,b⟩) a proper subgroup. Let ⟨a⟩ be a cyclic group of order k|A| where k is coprime to ⟨b⟩, and define ba = bA. Then the center of ⟨a,b⟩ is still a proper subgroup and contains ⟨a|A|⟩ of order k. More concretely:

Let b have odd order m and let a have order 2n+1 and let a act on b as ba = b−1. Then ⟨a,b⟩ has center ⟨a2⟩ of order 2n.

You can combine the two directions, but again because the first direction trivializes, the combined version looks pretty silly. Just take the direct product any example of the second direction and a cyclic group of order coprime to its |a|. This exhausts all possibilities, but I suspect most people would be irritated with the direct product aspect and only consider the second direction valid.

If you relax the coprime hypothesis, then you get another type of example that is reasonably interesting: take the Sylow p-subgroup of the holomorph of a cyclic p-group. The center will have order exactly p. If ⟨b⟩ has order pn+1, then ⟨a⟩ will have order pn, ⟨a,b⟩ will have order p2n+1, and the center of ⟨a,b⟩ will be ⟨bpn⟩ of order p. The action of a on b is easy to write down: ba = b1+p.