In the following let $X,Y,Z$ be independent standard normals in $\R^d$. Then for all $f,g\in L_2(\g)$:
$$
\int P_tf.g\,d\g
=\E(P_tf(X)g(X))
=\E(f(e^{-t}X+\sqrt{1-e^{-2t}}Y)g(X))
=\E(f(e^{-t}X-\sqrt{1-e^{-2t}}Y)g(X))~.
$$
The last equality is due to the fact that $(X,-Y)$ and $(X,Y)$ have the same distribution! Put $U=e^{-t}X+\sqrt{1-e^{-2t}}Y$ and $V=e^{-t}Y-\sqrt{1-e^{-2t}}X$. Since the transformation
$$
(x,y)\mapsto((e^{-t}x+\sqrt{1-e^{-2t}}y,e^{-t}y-\sqrt{1-e^{-2t}}x)
$$
is orthogonal on $\R^{2d}=\R^d\times\R^d$, the random variables $U,V$ are independent standard normals and $X=e^{-t}U-\sqrt{1-e^{-2t}}V$. It follows that
$$
\E(f(e^{-t}X+\sqrt{1-e^{-2t}}Y)g(X))
=\E(f(U)g(e^{-t}U-\sqrt{1-e^{-2t}}V))
=\E(f(U)P_tg(U))
$$
Thus $P_t:L_2(\g)\rar L_2(\g)$ is self-adjoint.
\begin{eqnarray*}
\E(P_tP_sf(X)g(X))
&=&\E(P_sf(e^{-t}X+\sqrt{1-e^{-2t}}Y)g(X))\\
&=&\E(f(e^{-s}(e^{-t}X+\sqrt{1-e^{-2t}}Y)+\sqrt{1-e^{-2s}}Z)g(X))\\
&=&\E(f(e^{-t-s}X+e^{-s}\sqrt{1-e^{-2t}}Y+\sqrt{1-e^{-2s}}Z)g(X))
\end{eqnarray*}
Since $(X,aY+bZ)$ has the same distribution as $(X,\sqrt{a^2+b^2}Y)$ we simply have to verify that
$$
e^{-2s}(1-e^{-2t})+(1-e^{-2s})=1-e^{-2s-2t}
$$
Hence $P_sP_t=P_{s+t}$. Finally for e.g. $f\in C_c^\infty(\R^n)$ we put $h=(e^{-t}-1)x+\sqrt{1-e^{-2t}}y$ and get:
$$
f(x+h)
=f(x)
+\la\nabla f(x),h\ra+\tfrac12\Hess f(x)(h,h)
+\Oh(\norm h^3)
$$
Therefore $(P_tf(x)-f(x))/t$ is given by
$$
\int\frac{\la\nabla f(x),h\ra+\tfrac12\Hess f(x)(h,h)}t\,\g(dy)+\Oh(\sqrt t)
$$
Now $\int\la\nabla f(x),y\ra\,\g(dy)=0$, $\int y_jy_k\,\g(dy)=\d_{jk}$ and
$$
\lim_{t\dar0}\frac h{\sqrt t}=y~.
$$
Hence we get for all $f\in C_c^\infty(\R^n)$:
$$
\lim_{t\dar0}\frac{P_tf(x)-f(x)}t
=\int-\la\nabla f(x),x\ra+\Hess f(x)(y,y)\,\g(dy)
=-Hf(x)~.
$$