The one line answer is that the category $\mathsf{Ab}$ of abelian groups is enriched over the skew-monoidal category $\mathsf{Gp}$ of groups, and that this "faux-tensor" defines a skew-action of the skew-monoidal category $\mathsf{Gp}$ on $\mathsf{Ab}$.
A skew-monoidal structure on a category $\mathcal{C}$ consists of a "tensor product" functor $\boxtimes \colon \mathcal{C} \times \mathcal{C} \to \mathcal{C}$, a "unit" object $I \in \mathcal{C}$, and "associativity and unit constraint" natural transformations $\alpha \colon (X \boxtimes Y) \boxtimes Z \to X \boxtimes (Y \boxtimes Z)$, $\lambda \colon I \boxtimes X \to X$, and $\rho \colon X \to X \boxtimes I$, satisfying the original five coherence axioms of Mac Lane. The important point is that these associativity and unit constraints are not required to be invertible. This notion was introduced by Szlachányi in his paper
Kornél Szlachányi. Skew-monoidal categories and bialgebroids. Adv. Math. 231 (2012), no. 3-4, 1694--1730. https://doi.org/10.1016/j.aim.2012.06.027
and has been much studied since, especially by the Australian school of category theory.
The "half-tensor products" of groups that you describe are part of a skew-monoidal structure on the category $\mathsf{Gp}$ of groups. This skew-monoidal structure is an instance of the family of examples described in Example 2.7 of my paper:
Alexander Campbell. Skew-enriched categories. Applied Categorical Structures 26 (2018), no. 3, 597--615. https://doi.org/10.1007/s10485-017-9504-0
The tensor product $G \boxtimes H$ of two groups $G$ and $H$ is the group you denote by $G \triangleleft H$, i.e. the copower of $G$ by the underlying set of $H$. Note that group homomorphisms $G \boxtimes H \to K$ correspond to functions $G \times H \to K$ that are group homomorphisms in the first variable. The unit object is the free group on one generator, i.e. $\mathbb{Z}$. The associativity and unit constraints are a little more complicated to describe, but suffice it to say that they are not invertible.
This skew-monoidal structure on $\mathsf{Gp}$ is closed: the functor $- \boxtimes H$ has a right adjoint which sends a group $K$ to the group $[H,K]$ of all functions from $H$ to $K$ with the pointwise group structure; this group $[H,K]$ is the internal hom for this skew-monoidal structure on $\mathsf{Gp}$. Thus $\mathsf{Gp}$ is also a skew-closed category in the sense introduced by Ross Street in his paper:
Ross Street. Skew-closed categories. J. Pure Appl. Algebra 217 (2013), no. 6, 973--988. https://doi.org/10.1016/j.jpaa.2012.09.020
Now, just as one can define categories enriched over monoidal categories, one can also define categories enriched over skew-monoidal categories. (In the terminology of my paper cited above, this is the same thing as a "left normal skew-enrichment" over the skew-monoidal category. Enrichment over skew-closed categories is defined in Street's paper cited above.)
We can define an enrichment of $\mathsf{Ab}$ over the above skew-monoidal structure on $\mathsf{Gp}$ as the change of base of the usual self-enrichment of $\mathsf{Ab}$ along the inclusion functor $\mathsf{Ab} \to \mathsf{Gp}$ equipped with the lax monoidal structure whose tensor constraint $A \boxtimes B \to A \otimes B$ is the homomorphism $U(B) \odot A \to A \otimes B$ whose component at an element $b \in B$ is $-\otimes b \colon A \to A \otimes B$.
Unpacking this, we have that, for each pair of abelian groups $A$ and $B$, the hom-group $\underline{\operatorname{Hom}}(A,B)$ is the usual group of group homomorphisms from $A$ to $B$, with its pointwise group structure, but where we have forgotten that it's abelian. For each triple of abelian groups $A$, $B$, and $C$, the composition homomorphism $\underline{\operatorname{Hom}}(B,C) \boxtimes \underline{\operatorname{Hom}}(A,B) \to \underline{\operatorname{Hom}}(A,C)$ corresponds to the usual composition function $\operatorname{Hom}(B,C) \times \operatorname{Hom}(A,B) \to \operatorname{Hom}(A,C)$, but where we have forgetten that it's a group homomorphism in the second variable. Similarly, the unit homomorphisms $\mathbb{Z} \to \underline{\operatorname{Hom}}(A,A)$ simply pick out the identity homomomorphisms.
(Note that this enrichment of $\mathsf{Ab}$ over $\mathsf{Gp}$ can also be seen an instance of Example 2.7 of my paper cited above, since the category of abelian groups is equivalent to the category of group objects in $\mathsf{Gp}$.)
As you've spelled out in your question, the hom-functor $\underline{\operatorname{Hom}} \colon \mathsf{Ab}^\mathrm{op} \times \mathsf{Ab} \to \mathsf{Gp}$ is part of a two-variable adjunction, and so there are defined tensoring and cotensoring operations of an abelian group by a group. In particular, the tensoring operation defines a skew-action of the skew-monoidal category $\mathsf{Gp}$ on the category $\mathsf{Ab}$, in the sense of the paper:
Stephen Lack and Ross Street. Skew-monoidal reflection and lifting theorems. Theory Appl. Categ. 30 (2015), Paper No. 28, 985--1000. http://tac.mta.ca/tac/volumes/30/28/30-28abs.html
Note that a skew-action of a skew-monoidal category $\mathcal{V}$ on a category $\mathcal{C}$ is simply an oplax monoidal functor $\mathcal{V} \to \operatorname{Fun}(\mathcal{C},\mathcal{C})$.
Best Answer
The answer is yes if you assume enough things. In particular, the notion of a left flat object of $\mathcal A$ comes up :
Definition: An object $L\in\mathcal A$ is left flat if $-\otimes L$ is exact.
My assumption will be that $\mathcal A$ has enough left flat objects. This allows you to even define $\mathbb L(P\otimes_Q -)$ in terms of left flat resolutions as I will explain below.
(Note : Here I'm using a definition of $\mathbb L$ which is slightly more general than "take projective resolutions", namely I'm using the one using "left defomations", see Riehl's book on homotopical algebra. If you want to use projective resolutions, the assumption is that $\mathcal A$ has enough left flat projectives)
Warning: Note that if $\mathcal A$ has projectives that are not left flat, then the answer is no even for $Q=*$, as is easy to convince yourself of. In particular, if $\mathcal A$ has enough projectives (to define $\mathbb L$ in the classical homological algebra sense), the assumption is that enough (equivalently all) projectives are left flat.
In this case, the answer is yes, and basically the proof is the same as in the ordinary case. Let me work in the bounded below case because there are the same subtleties as in the ordinary case for the unbounded case.
In particular, up to shifting, I will work in the connective case.
Let me sketch a proof below (convention : my functor categories are categories of $Ab$-enriched functors, I'm assuming this is what you meant - otherwise, replace $\hom_Q$ with $\mathbb Z[\hom_Q]$):
Definition: $P$ is right flat if $P\otimes_Q -$ is exact, and same for $X$ being left flat.
Lemma: Projectives in $Fun(Q^{op},Ab)$ are right flat, and there are enough left flats in $Fun(Q,\mathcal A)$.
Proof: By the enriched Yoneda lemma and smallness of $Q$, every projective is a summand of $\bigoplus_i \hom_Q(-,q_i)$ for some family of $q_i$'s. Therefore it suffices to prove it for those ones, because flatness is stable under retracts. Flatness is also stable under direct sums because they are exact ($\mathcal A$ is Grothendieck, in particular AB5), so it suffices to prove it for $\hom_Q(-,q)$. But now by the enriched Yoneda lemma again (the "canonical colimit of representable presheaves" version), we have $\hom_Q(-,q)\otimes_Q X \cong X(q)$, which is manifestly an exact functor.
The dual case is dual, you simply need to allow $\hom_Q(q,-)\otimes L$ for enough flat objects $L$ of $\mathcal A$.
In particular this lemma tells you that the two things you want to compare are well-defined.
Corollary: If $C$ is a chain complex of right flat functors $Q^{op}\to Ab$, $C\otimes_Q -$ preserves quasi-isomorphisms (where you define $\otimes_Q$ on chain complexes in the obvious way); and dually.
Proof : Let $n\in \mathbb N$, and let $C_{\leq n}$ denote the so-called stupid truncation of $C$, i.e. $0\to C_n\to ... \to C_0$. Because $-\otimes_Q -$ manifestly preserves filtered colimits in each variable, because filtered colimits are exact ($\mathcal A$ is Grothendieck, in particular AB5), and finally because $C= \mathrm{colim}_n C_{\leq n}$, it suffices to prove that each $C_{\leq n}\otimes_Q -$ preserves quasi-isomorphisms.
Now we have a short exact sequence $0\to C_{\leq n-1}\to C_{\leq n}\to C_n\to 0$ which is split as a sequence of graded objects. Note that the underlying graded object of $A\otimes_Q B$ only depends on the underlying graded objects of $A,B$ respectively. It follows that this exact sequence remains exact after tensoring (over $Q$) with anything.
In particular, by the long exact sequence in homology, and by induction, we reduce to proving that $C_n\otimes_Q-$ preserves quasi-isomorphisms for each $n$. But this is by the assumption that each $C_n$ is right flat.
The dual case is completely dual.
Note : I am implicitly using the fact that $-\otimes -: Ab\times\mathcal A\to\mathcal A$ preserves colimits in each variable. This is an exercise I'll leave to you :)
Note : This lemma is essentially what allows you to define $\mathbb L$ using left/right flat resolutions.
Note: This lemma is often packaged as a spectral sequence argument, but it's really about the filtration.
Corollary: $\mathbb L(P\otimes_Q -)(X)\simeq \mathbb L(-\otimes_Q X)(P)$.
Proof : Pick a right flat resolution $\tilde P$ of $P$, and a left flat resolution $\tilde X$ of $X$ (those exist by the first lemma).
You have a zigzag $$\mathbb L(P\otimes_Q -)(X) =P\otimes_Q \tilde X \to \tilde P\otimes_Q \tilde X \leftarrow \tilde P\otimes_Q X = \mathbb L(-\otimes_Q X)(\tilde P)$$
The extreme equalities are by definition, and the two middle arrows are quasi-isomorphisms by the previous corollary.
Note : this quasi-isomorphism can be made as natural as left flat resolutions in $\mathcal A$.