You probably know this, but your conditions can't be met by any principal $p$-block of any finite group $G$ which is not $p$-solvable, although I think the Classification of Finite Simple Groups (CFSG) is necessary for that. For suppose that $G$ is a finite group with cyclic Sylow $p$-subgroup $D$ and with $|D| >p,$ but that $G$ is not $p$-solvable. Then the principal $p$-block $B$ of $G$ has defect group $D$, and this does not change on passage to $G/O_{p^{\prime}}(G),$ so we might as well suppose that $O_{p^{\prime}}(G) = 1.$ Now $O_{p}(G) = 1,$ for otherwise we have $D_{1} \lhd G,$ and then $C_{G}(D_{1}) \lhd G.$ But $C_{G}(D_{1})$ has a (characteristic) normal $p$-complement ( which must be trivial) as $O_{p^{\prime}}(G) = 1.$ Then $C_{G}(D_{1}) = D \lhd G,$ and $G$ is certainly $p$-solvable, a contradiction.

Let $H$ be a Hall $p^{\prime}$-subgroup of $N_{G}(D).$ Then $D = [D,H] \times C_{D}(H)$ since $D$ is Abelian of order coprime to $|H|$. Since $D$ is cyclic, we either have $C_{D}(H) = D$ or $C_{D}(H) = 1.$ In the former case, $N_{G}(D)$ has a normal $p$-complement, and then so does $G,$ contrary to the fact that $G$ is not $p$-solvable. Hence $D = [D,H] \leq G^{\prime}.$

Now $G^{\prime}$ is not $p$-solvable, as $G$ is not, so we then obtain $D \leq G^{\prime \prime}$ by the same argument, and ultimately $D \leq G^{(\infty)},$ the terminal member of the derived series for $G$, by repeating the argument.

Let $M$ be a minimal normal subgroup of $G.$ Then $M$ has order divisible by $p$, so that $D_{1} \leq M.$ Also $M$ is not a $p$-group, so $M$ is non-Abelian simple (using the fact that $M$ has cyclic Sylow $p$-subgroup). Then $G = MN_{G}(D_{1})$ by a Frattini (type) argument.Now $MC_{G}(D_{1}) \lhd G$ and $G/MC_{G}(D_{1})$ is a homomorphic image of the Abelian $p^{\prime}$-group $N_{G}(D_{1})/C_{G}(D_{1}),$ so is itself Abelian of order prime to $p$.

Now we have $D \leq MC_{G}(D_{1})$ and $MC_{G}(D_{1})/M$ (being a homomorphic image of $C_{G}(D_{1})$ ) has a normal $p$-complement. But $MC_{G}(D_{1})$ is certainly not $p$-solvable, so arguing as before, we have $D \leq [MC_{G}(D_{1})]^{\prime}.$ Hence $MC_{G}(D_{1})$ has no factor group of order $p$, so that $MC_{G}(D_{1})/M$ is a $p^{\prime}$-group. Now $G/M$ is a $p^{\prime}$-group, so that $D \leq M.$

Now $G = MN_{G}(D)$ by a Frattini argument. I claim that we now have $N_{G}(D) = N_{G}(D_{1}).$ This is the (first and only) time we require CFSG, though the previous argument could have been shortened considerably using CFSG.

Note that $N_{G}(D_{1}) = N_{M}(D_{1})N_{G}(D),$ so to prove the claim, it suffices to prove that $N_{M}(D_{1}) = N_{M}(D).$ But it is a Theorem of H. Blau that whenever a finite non-Abelian simple group $X$ has a cyclic Sylow $p$-subgroup $P \neq 1$, then $P$ is TI in $X$, that is, $P \cap P^{x} = 1$ for all $x \in X \backslash N_{X}(P)$.

In our situation, this immediately yields that $N_{M}(D_{1}) = N_{M}(D),$ as required. Hence your fourth bulletpoint can't be satisfied (for principal blocks of non-$p$-solvable groups) as remarked in comments.