Given a function $g \in C^2(S^{n-1})$ the unit sphere then the unique solution to laplaces equation on the unit ball with this boundary data, is given by reflecting the green's function for the halfspace, through the ball, and is $u(x)=\frac{1-|x|^2}{n \alpha(n)} \int g(y)/|x-y|^n dS(y)$. I.e. the green's function $G(x,y)$ is $\frac{1-|x|^2}{n \alpha(n)} *1/|x-y|^n$. This works for $n\geq 2$. Note that this works for $n=2$
Now in dimension $n=2$, if you express $u$ in polar coordinates, then $u$ can be written as a convolution with an approximate identity $P_{r,\theta}$(the poisson kernel). This is an approximate identity in the sense that it has integral in $\theta$ equal to 1 and , its $L^1$ norm which is just its integral is bounded(=1) as $r$ tends to 1 and $\int_{\delta < \theta < 2\pi -\delta} P_{r,\theta}$ tends to 0 as $r \to 1$. I.e. it is the same thing as your usual approximate identity with $N \to \infty$ replaced by $r \to 1$.
It should be possible for to write the solution to laplace's equation in higher dimensions as a convolution with an approximate identity. This is because the representation formula approaches the boundary data as you approach the boundary.
In light of this, I want to ask a question: Is there a way that $G(x,y)$ can be written as an approximate identity in any sense?
Strictly speaking, a convolution ($\int f(x-y)g(y)\,dy$) makes sense only on an abelian group. In particular, on $S^1$ which is an abelian group. But for $n>2$, the sphere $S^{n-1}$ does not have a reasonable structure of an abelian group ($S^3$ is the group of unit quaternions, and others aren't any kind of a group). Thus, the meaning of approximate identity won't be quite the same in higher dimensions. There is no operation for which to be an identity...
But we can still talk about the Dirac delta $\delta[f] = f(e_1)$, which evaluates functions on $S^{n-1}$ at the point $e_1$ (first basis vector). Setting $x = re_1$ and let $r\to 1$, you get a family of functions $P_r(y) = P(re_1,y)$ on $S^{n-1}$ that converge to $\delta$ weakly. Indeed, for any continuous function $f$ on $S^{n-1}$ the Poisson integral $F$ is continuous on $B^n$, hence $F(re_1)\to f(e_1)$ as $r\to 1$.
You can express the values of the Poisson integral on the sphere of radius $r$ by a convolution-type integral involving $P_r(y)$, taken with respect to Haar measure $\nu$ on the special orthogonal group $SO(n)$: $$ u(rx) = \int_{SO(n)} f(x) P_r(Ax)\,d\nu(A) $$
Integration over the Haar measure on $SO(n)$ is a substitute for group structure on $S^{n-1}$.
I don't see where Green's function fits here. A convolution-type integral with Green's function, $$ \int_{B^n } f(y) G(x,y)\,dy $$ solve the Poisson equation in $B^n$ with homogeneous boundary condition, not the Laplace equation with given boundary condition. Of course $G$ and $P$ are related. But it's $P$ that looks like an identity.