Let me specialize heavily to the case of formal groups (group laws) of dimension one over a $p$-adic ring $\mathfrak o$, i.e. the ring of integers of a finite extension $k$ of $\Bbb Q_p$.

I still am uncertain about what category you’re thinking of. If we restrict further to formal groups of finite height (the endomorphism $[p]$ being of finite degree $p^h$), then these things become $p$-divisible groups, or, if you like, ind-finite objects. For instance the kernel of $[p^n]$ will be a finite $\mathfrak o$-group-scheme, $K_n=\ker([p^n])=\mathrm{Spec}(\mathfrak o[[x]]/([p^n](x))\,)$, and you have natural maps $K_n\hookrightarrow K_{n+1}$, and you see that $\projlim\mathfrak o[[x]]/([p^n](x))\cong\mathfrak o[[x]]$. In this sense, your $G$, if indeed a formal group of finite height over $\mathfrak o$, is the union of its finite subgroups. This is the viewpoint that I tend to work with.

Now, let’s consider just one fairly simple case, where the formal group law has all its coefficients in an unramified extension $A$ of $\Bbb Z_p$, even in $\Bbb Z_p$ itself, and suppose the height is $h=2$ for simplicity. This means that $[p](x)\equiv px+ux^{p^2}\pmod{x^{p^2+1}}$, where $u$ is a unit of $A$, and the congruence ignores all terms in the power series of degree $>p^2$. Look at the Newton polygon and see that all the $z\in\overline k$ with $v_p(z)>0$ and $[p](z)=0$ have $v_p(z)=\frac1{p^2-1}$, plus of course $0$. So $p^2$ in all, and thus they form an elementary $p$-group of order $p^2$.

Now take any of the cyclic subgroups of $\ker[p]$, call it $\Gamma$. One proves that $$ \pi_\Gamma(x)=\prod_{\gamma\in\Gamma}F(x,\gamma)\,, $$ which is defined over a totally ramified extension $A'$ of $A$ (actually of degree $p+1$), is a morphism into another formal group, which I will abuse language in calling $G/\Gamma$.

I ask you to believe that I have shown you a formal group $G/\Gamma$ that, as far as I can see, will prove to you that $\Gamma$ is not a special subgroup of $G$, once you see that the formal group law of $G/\Gamma$ is not isomorphic to that of $G$, not even with a morphism $\varphi^*$ of the type you allow. (I think, because I’m not sure what properties you allow $\varphi^*$ to have.)

How do I know that $G/\Gamma$ is nothing like $G$? By Newtonian magic, you see that the Newton polygon of $[p]_{G/\Gamma}$ has vertices at $(1,1)$, $(p,\frac1{p+1})$, and $(p^2,0)$. The important fact is that this polygon is not the same as that of $[p]_G$; and since the shape of the Newton polygon of $[p]$ is an invariant, it follows that there is no way for $G/\Gamma$ to be isomorphic to $G$.

In the appropriate category, the map from $G$ to $G/\Gamma$ is onto. You can show, for instance, that if $v_p(\eta)>0$, there is $\xi$ in a finite extension of $k(\eta)$ such that $v_p(\xi)>0$ and $\pi_\Gamma(\xi)=\eta$.

(All of this is in an old and poorly-written paper of mine, Finite subgroups and isogenies…)

Let me specialize heavily to the case of formal groups (group laws) of dimension one over a $p$-adic ring $\mathfrak o$, i.e. the ring of integers of a finite extension $k$ of $\Bbb Q_p$.

I still am uncertain about what category you’re thinking of. If we restrict further to formal groups of finite height (the endomorphism $[p]$ being of finite degree $p^h$), then these things become $p$-divisible groups, or, if you like, ind-finite objects. For instance the kernel of $[p^n]$ will be a finite $\mathfrak o$-group-scheme, $K_n=\ker([p^n])=\mathrm{Spec}(\mathfrak o[[x]]/([p^n](x))\,)$, and you have natural maps $K_n\hookrightarrow K_{n+1}$, and you see that $\projlim\mathfrak o[[x]]/([p^n](x))\cong\mathfrak o[[x]]$. In this sense, your $G$, if indeed a formal group of finite height over $\mathfrak o$, is the union of its finite subgroups. This is the viewpoint that I tend to work with.

Now, let’s consider just one fairly simple case, where the formal group law has all its coefficients in an unramified extension $A$ of $\Bbb Z_p$, even in $\Bbb Z_p$ itself, and suppose the height is $h=2$ for simplicity. This means that $[p](x)\equiv px+ux^{p^2}\pmod{x^{p^2+1}}$, where $u$ is a unit of $A$, and the congruence ignores all terms in the power series of degree $>p^2$. Look at the Newton polygon and see that all the $z\in\overline k$ with $v_p(z)>0$ and $[p](z)=0$ have $v_p(z)=\frac1{p^2-1}$, plus of course $0$. So $p^2$ in all, and thus they form an elementary $p$-group of order $p^2$.

Now take any of the cyclic subgroups of $\ker[p]$, call it $\Gamma$. One proves that $$ \pi_\Gamma(x)=\prod_{\gamma\in\Gamma}F(x,\gamma)\,, $$ which is defined over a totally ramified extension $A'$ of $A$ (actually of degree $p+1$), is a morphism into another formal group, which I will abuse language in calling $G/\Gamma$.

I ask you to believe that I have shown you a formal group $G/\Gamma$ that, as far as I can see, will prove to you that $\Gamma$ is not a special subgroup of $G$, once you see that the formal group law of $G/\Gamma$ is not isomorphic to that of $G$, not even with a morphism $\varphi^*$ of the type you allow. (I think, because I’m not sure what properties you allow $\varphi^*$ to have.)

How do I know that $G/\Gamma$ is nothing like $G$? By Newtonian magic, you see that the Newton polygon of $[p]_{G/\Gamma}$ has vertices at $(1,1)$, $(p,\frac1{p+1})$, and $(p^2,0)$. The important fact is that this polygon is not the same as that of $[p]_G$; and since the shape of the Newton polygon of $[p]$ is an invariant, it follows that there is no way for $G/\Gamma$ to be isomorphic to $G$.

In the appropriate category, the map from $G$ to $G/\Gamma$ is onto. You can show, for instance, that if $v_p(\eta)>0$, there is $\xi$ in a finite extension of $k(\eta)$ such that $v_p(\xi)>0$ and $\pi_\Gamma(\xi)=\eta$.

(All of this is in an old and poorly-written paper of mine, Finite subgroups and isogenies…. EDIT:The “Newtonian magic” involves the “Newton copolygon”, also called the valuation function. I’ll bet a nickel that somebody else has explained it better than I can, but it’s in a later paper of mine, Canonical subgroups of formal groups, and I fear that it’s at most a little better-written than the other. Copolygon talk begins on p. 109.)