The answer is no: every such full subcategory is stable under direct products, or equivalently $L$ preserves direct products.
Here all Hom are in the category of groups. The definition of $L$ means that $L$ is a covariant functor and for every group there is a group homomorphism $G\to LG$, so that the obvious squares commute, and such that the natural map $\mathrm{Hom}(LG,U)\to \mathrm{Hom}(G,U)$ induced by $G\to LG$ is a bijection for all $U\in D$.
Idea of the proof. Categorically usually describes $G\times H$ by showing that it represents $K\mapsto\mathrm{Hom}(K,G)\times \mathrm{Hom}(K,H)$. But one needs here to describe it on the left, i.e. to describe $\mathrm{Hom}(G\times H,K)$, which can be described as the set of pairs $(u,v)$ in $\mathrm{Hom}(H,K)\times\mathrm{Hom}(G,K)$ whose images centralize each other.
Next, one observe that for $U\in D$, both $\mathrm{Hom}(L(G\times H),U)$ and $\mathrm{Hom}(LG\times LH,U)$ can be described as the set of pairs in $\mathrm{Hom}(H,K)\times\mathrm{Hom}(G,K)$ whose images centralize each other; more precisely, we have a canonical homomorphism $\pi:L(G\times H)\to LG\times LH$ inducing a bijection $\mathrm{Hom}(LG\times LH,U)$ and $\mathrm{Hom}(L(G\times H),U)$ for all $U\in D$ (given by $f\mapsto f\circ \pi$). One would like to conclude from Yoneda's lemma; however we cannot because $LG\times LH$ might not be in $D$. Nevertheless, we can argue as in the proof of Yoneda's lemma, picking $U=L(G\times H)$ and consider a preimage of the identity map of $L(G\times H)$. This yields $f$ such that $f\circ\pi$ is the identity of $L(G\times H)$ and thus $\pi$ is injective.
So to conclude we need to show that $\pi$ is surjective and this is actually the beginning of the proof below, since it uses very little.
Here's now the detailed proof.
First observe that $L$ maps the trivial group 1 to itself (edit: alternatively, see the comment by Todd Trimble). Indeed we have a bijection (the previous one for $G=1$, $U=L1$) $\mathrm{Hom}(L1,L1)\to \mathrm{Hom}(1,L1)$; the latter is a singleton and the former contains the identity of $L1$ and the trivial endomorphism and hence is a singleton iff $L1=1$. It follows that $L$ maps the trivial (=constant) homomorphisms to themselves (because these are the one factoring through the trivial group).
For any groups $G,H$ use the notation for the canonical injections and projections:
$$ G\stackrel{i_G}\to G\times H\stackrel{i_H}\leftarrow H,\quad G\stackrel{p_G}\leftarrow G\times H\stackrel{p_H}\to H,$$
inducing homomorphisms
$$ LG\stackrel{Li_G}\to L(G\times H)\stackrel{Li_H}\leftarrow LH,\quad LG\stackrel{Lp_G}\leftarrow L(G\times H)\stackrel{Lp_H}\to LH.$$
We have
$$p_Gi_G=\mathrm{id}_G,\quad p_Gi_G=\mathrm{id}_G, \quad p_Gi_H=1, \quad p_Hi_G=1,$$
thus
$$Lp_GLi_G=\mathrm{id}_{LG},\quad Lp_HLi_H=\mathrm{id}_{LH}, \quad Lp_GLi_H=1_{LH,LG}, \quad Lp_HLi_G=1_{LG,LH},$$
where $1_{A,B}$ is the trivial (constant) homomorphism from $A$ to $B$. From the first two equalities, we deduce that $Li_G$ and $Li_H$ are injective and we have semidirect decompositions
$$L(G\times H)=Li_G(LG)\ltimes \mathrm{ker}(Lp_G)=Li_H(LH)\ltimes \mathrm{ker}(Lp_H)$$
(where the projection to the left-hand factor is given by $Li_GLp_G$, respectively $Li_HLp_H$);
from the last two equalities we deduce that $Li_G(LG)\subset \mathrm{ker}(Lp_H)$ and $Li_H(L_H)\subset \mathrm{ker}(Lp_G)$. It follows that the map
$$L(G\times H)\stackrel{Lp_G\times Lp_H}\to LG\times LH$$
is surjective: indeed the left-hand map being surjective in restriction to $Li_G(LG)$, it amounts to see that the right-hand map is surjective in restriction to the kernel of the left-hand map, and this kernel is $\mathrm{ker}(Lp_G)$ (by the first semidirect decomposition), and this kernel contains $Li_H(L_H)$, in restriction to which the right-hand map is surjective. This proves that $Lp_G\times Lp_H$ is surjective (I used very little of the hypotheses about $L$: only that it's an endofunctor mapping the trivial groups to themselves).
Now we proceed to show that $Lp_G\times Lp_H$ is also injective.
If $f:A\to B$ is a group homomorphism and $K$ another group, denote by $q_K[f]$ the resulting map $\mathrm{Hom}(B,K)\to\mathrm{Hom}(A,K)$ given by $g\mapsto g\circ f$.
Consider the following maps:
$$[*]\quad \mathrm{Hom}(LG\times LH,K)\stackrel{q_K[Lp_G\times Lp_H]}\to \mathrm{Hom}(L(G\times H),K)\stackrel{q_K[Li_G]\times q_K[Li_H]}\to \mathrm{Hom}(LG,K)\times\mathrm{Hom}(LH,K).$$
Let $s$ be the composite map. Let us show that $s$ is injective and describe its image.
Since $q_K$ is a contravariant functor, the composite map $s$ is given by $$s=q_K[(Lp_G\times Lp_H)\circ Li_G]\times q_K[(Lp_G\times Lp_H)\circ Li_H]$$
$$=q_K[Lp_GLi_G\times Lp_HLi_G]\times q_K[Lp_GLi_H\times Lp_HLi_H]$$
$$=q_K[\mathrm{id}_{LG}\times 1_{LG,LH}]\times q_K[1_{LH,LG}\times \mathrm{id}_{LH}].$$
Explicitly, $\mathrm{Hom}(LG\times LH,K)$ is given (via the embedding $s$) by the set of pairs of homomorphisms $(u_G,u_H)\in\mathrm{Hom}(LG,K)\times\mathrm{Hom}(LH,K)$ whose images in $K$ centralize each other; here if $u\in \mathrm{Hom}(LG\times LH,K)$, $u_G$ is its restriction to $G$ and $u_H$ its restriction to $H$, and $u(x,y)=u_G(x)u_H(y)$ for all $(x,y)\in LG\times LH$. We see (from the above cumbersome but trivial computation) that $s(u)=(u_G,u_H)$.
Now assume that $K=U$ belongs to $D$. Then the map
$$\mathrm{Hom}(L(G\times H),U)\stackrel{q_K[Li_G]\times q_K[Li_H]}\to \mathrm{Hom}(LG,U)\times\mathrm{Hom}(LH,U)$$
is isomorphic (by the functorial property of $L$) to the map
$$\mathrm{Hom}(G\times H,U)\stackrel{q_K[i_G]\times q_K[i_H]}\to \mathrm{Hom}(G,U)\times\mathrm{Hom}(H,U),$$
which itself is injective, for the same reason as the previous one (a homomorphism on a direct product is determined by its restriction to factors) and also, for the same reason its image consists of pairs of homomorphisms whose images centralize each other.
So, writing symbolically [$*$] as $X\stackrel{v}\to Y\stackrel{w}\to Z$ we have shown that $w\circ v$ and $w$ are both injective with the same image (**edit: [$**$]). This implies that $v$ is bijective. Namely, for every $U$ in $D$,
$$\mathrm{Hom}(LG\times LH,U)\stackrel{q_K[Lp_G\times Lp_H]}\to \mathrm{Hom}(L(G\times H),U)$$
is bijective.
Apply this (the surjectivity) to $U=L(G\times H)$ and the identity of $U$, we obtain that there exists a homomorphism $g:LG\times LH\to L(G\times H)$ such that $g\circ (Lp_G\times Lp_H)=\mathrm{id}_{L(G\times H)}$. This shows that $Lp_G\times Lp_H$ is injective and concludes the proof.
Edit: in [$**$] I have to justify that they have the same images. I was initially not careful about this: if $G\to LG$ is surjective for all $G$, then this is fine because the above description really provides the same description for pairs of homomorphisms extending to a homomorphism on the direct product. But if $G\to LG$ is not surjective, it's a priori not clear that if we have $G\times H\to U$, then the images of the factored maps $LG\to U$ and $LH\to U$ centralize each other. However this is true by the lemma below, and thus it's fine.
Lemma If $G,U$ are groups with $U\in D$ and $f:G\to U$ induces $f':LG\to U$, then $C_U(\mathrm{Im}(f))=C_U(\mathrm{Im}(f'))$, where $C_U$ denotes the centralizer in $U$. In particular, if $g:H\to U$ is another such homomorphism and induces $g'$ and if the images of $f$ and $g$ centralize each other, then the images of $f'$ and $g'$ centralize each other.
Proof: first note that $\mathrm{Im}(f)\subset\mathrm{Im}(f')$, so the inclusion $C_U(\mathrm{Im}(f))\supset C_U(\mathrm{Im}(f'))$ is clear. Now consider $t\in C_U(\mathrm{Im}(f))$. Then $h:y\mapsto tf'(y)t^{-1}$ is a homomorphism from $LG$ to $U$ and the composite map $G\to U$ equals $f$. By the universal property of $L$ (namely, the injectivity of $\mathrm{Hom}(LG,U)\to\mathrm{Hom}(G,U)$), it follows that $h=f'$. So $t$ centralizes the image $\mathrm{Im}(f')$. This proves the first statement.
The second statement follows by applying the first statement twice: if $u$ belongs to the image of $f$, then it centralizes the image of $g$, hence centralizes the image of $g'$. Now if $u'$ belongs to the image of $g'$, then it centralizes the image of $f$, hence centralizes the image of $f'$.
