Let $H$ denote $SL_2$.
By strong approximation, see http://www.jstor.org/stable/1970924, $H(K_\infty)G(K)$ is dense in $H({\mathbb A})$.
Now $H({\mathbb A}(\infty))$ contains $H(K_\infty)$ and contains a unit-neighborhood.
Therefore we have $H({\mathbb A})=H({\mathbb A}(\infty))H(K)$.
We can conclude the same thing for $G$, if we can show that $H(R)$ surjects onto $G(R)$ and likewise for $R={\mathbb A}, {\mathbb A}(\infty), K$.
The adelic result can be deduced from the result for fields and for $R={\cal O}_p$.

By Hilbert's Theorem 90, for each field $k$ the Galois cohomology $H^1(k,GL_1)$ is trivial and so the exact sequence of Galois cohomology shows that the sequence
$$
1\to GL_1(k)\to GL_2(k)\to PGL_2(k)\to 1
$$
is exact, which implies the claim for fields.
To verify the claim for $R={\cal O}_p$, we have to analyze the coordinate ring of $PGL_2$.
First, the coordinate ring of $GL_2$ over $R$ is
$$
A_{GL_2}= R[x_1,x_2,x_3,x_4,y]/(x_1x_4-x_2x_3)y-1.
$$
The coordinate ring of $PGL_2=GL_2/ GL_1$ is the ring of $GL_1$-invariants, where the action of $GL_1$ is given by
$$
\lambda.f(x_1,x_2,x_3,x_4,y)=f(\lambda x_1,\lambda x_2,\lambda x_3,\lambda x_4,\lambda^{-2} y).
$$
The ring of invariants is generated by all monomials of the form $x_ix_jy$ for $1\le i,j\le 4$.
Let now $\chi\in PGL_2({\cal O}_p)$.

Then $\chi$ is a homomorphism from $A_{PGL_2}$ to ${\cal O}_p$.

Every such can be extended to $A_{GL_2}\to K_p$ and we have to show that there exists an extension mapping to ${\cal O}_p$.
Pick any extension and denote it by the same letter $\chi$.
For the valuation $v$ on $K_p$ we have
$$
0\ \le\ v(\chi(x_j^2y))=2v(\chi(x_j))+v(\chi(y)).
$$
We are free to change $\chi(y)$ to $\chi(y)\pi^{2k}$ for any $k\in{\mathbb Z}$ if at the same time we change $\chi(x_j)$ to $\chi(x_j)\pi^{-k}$ and $\pi$ is a local uniformizer.
Thus we can assume $v(\chi(y))\in\{ 0,1\}$.
Then we conclude $v(\chi(x_j))\ge 0$ for every $j$ and so $\chi$ indeed maps into ${\cal O}_p$ as claimed.
This shows the result for ${\cal O}_p$.

Now let's put things together. We habe ${\mathbb A}^\times={\mathbb A}(\infty)^\times K^\times$, which is due to the fact that we are dealing with the curve ${\mathbb P}^1$ which has genus zero. Let $g\in GL_1({\mathbb A})$ then
$$
g=d(x,1)y,
$$
where $d(a,b)$ is the diagonal matrix with entries $a$ and $b$, $x$ is an idele and $y$ is in $SL_2({\mathbb A})$. By what we have shown, we have
$$
g=d(x_\infty x_K,1)y_\infty y_k=d(x_\infty,1)\tilde y d(x_K,1)y_K,
$$
where The $\infty$ indicates entries in ${\mathbb A}(\infty)$ and The $K$ indicates entries in $K$. The element $\tilde y$ is conjugate to $y_\infty$, therefore it no longer has entries in ${\mathbb A}(\infty)$, but it still lies in $SL_2$. Therefore, it can be decomposed again and finally we get $g=g_\infty g_K$.

We find that $h(G)$ is one. This however changes, if you take $K$ to be the rational function field of an arbitrary curve and ${\mathbb A}(\infty)$ the adeles which are unrestricted only at a given point.