As others have said, this is precisely the structure of a right module over the operad $E(D^p)$.

Since it doesn't feel right to give such a short answer that's already in the comments, let me expand a bit; I don't claim I'm saying anything new, but hopefully, maybe some readers will find something interesting.

If $P = \{P(k)\}_{k \ge 0}$ is an operad, a right $P$-module is given by a symmetric sequence $M$ equipped with composition maps
$$M(k) \otimes P(r_1) \otimes \dots \otimes P(r_k) \to M(r_1 + \dots + r_k)$$
satisfying the obvious axioms. If you define the plethysm of symmetric sequences $X \circ Y = \bigoplus_{i \ge 0} X(i) \otimes_{\mathfrak{S}_i} Y^{\otimes i}$, then an operad is a monoid wrt plethysm, and a right module is a right module over this monoid.

Since operads have units, this is equivalent to giving composition maps $\circ_i : M(k) \otimes P(l) \to P(k+l-1)$ for $1 \le i \le k$ (simply put, $m \circ_i p = m(1,\dots,1,p,1,\dots,1)$ where $1$ is the operadic unit and $p$ is in position $i$).

Plethysm is only linear on the left, so this actually gives two different notions of "left" stuff. On the one hand, a left $P$-module is a symmetric sequence $M$ equipped with composition maps:
$$P(k) \otimes M(r_1) \otimes \dots \otimes M(r_k) \to M(r_1 + \dots r_k).$$
This is exactly a left module over the monoid $P$ in the category of symmetric sequences and plethysms. Note that a left $P$-module concentrated in arity $0$ is the same thing as a $P$-algebra. On the other hand, you can define an "infinitesimal" or "abelian" left $P$-module, by only requiring composition maps $\circ_i : P(k) \otimes M(l) \to M(k+l-1)$. Since you don't have a unit in $M$, this is a different notion. (To be entirely honest, I do not know if the terminology "abelian/infinitesimal left module" is used. I am more used to the terminology "abelian/infinitesimal **bi**module", where you have a left and a right action defined as above.)

The precise modules in your question are very interesting, too. First, the operad $E(D^p)$ is weakly equivalent to the semi-direct product $E_p \rtimes O(p)$ of the little $p$-disks operad with the orthogonal group $O(p)$. If you require your embeddings to be oriented, you get a new operad $E^{or}(D^p)$ which is weakly equivalent to the usual framed little $p$-disks operad $E_p \rtimes SO(p)$. If you require your embeddings to preserve the canonical framing of $D^p$, then you get the usual little $p$-disks operad $E_p$.

If you have a smooth manifold $P$, then you can define the right $E(D^p)$-module $E(P)$ as you did in your question. Suppose you have another manifold $P'$ of dimension $p' \ge p + 3$. You can still define a right $E(D^p)$-module structure on $E(P')$, by considering embeddings of $D^p$ in $P'$.

Then Goodwillie--Weiss manifold calculus (following Arone, Boavida, Turchin, Weiss...) shows that the space of embeddings $\operatorname{Emb}(P,P')$ is weakly homotopy equivalent to the derived mapping space:
$$\operatorname{hMap}_{E(D^p)\text{-RMod}}(E(P), E(P')).$$
Here, ones takes e.g. a simplicial cofibrant replacement of $E(P)$ viewed as a right $E(D^p)$-module to obtain a simplicial set, given degree-wise by morphisms of right modules.

If $P$ is oriented then you can get away with using $E^{or}(D^p)$ instead, and if $P$ is parallelized you can even use the little $p$-disks operad instead. If the codimension $\dim P' - \dim P$ is less than $3$, then you obtain the analytic approximation $T_\infty \operatorname{Emb}(P,P')$ instead of the space of embeddings.