As requested, I'll elaborate here on the Gaussian case.

Suppose for instance that $X$ is a separable Banach space, and $\mathbb{P}$ is a Gaussian measure. Then each $f \in X^*$, when considered as a random variable on $(X, \mathbb{P})$ has a Gaussian distribution. Let $H$ denote the $L^2$-closure of $i X^*$. An $L^2$ limit of Gaussian random variables remains Gaussian, so every element of $H$ (and hence also $C \oplus H$) is Gaussian. But $L^2(X,\mathbb{P})$ clearly contains lots of non-Gaussian random variables (even when $X = \mathbb{R}^n$); as just one example, those of the form $f^2$ have a $\chi^2$ distribution.

However, since each $f \in X^*$ is Gaussian and hence has all moments, then for any polynomial $p$ in any number of variables $n$, we have that $p(f_1, \dots, f_n) \in L^2$. Then it can be shown by a Stone-Weierstrass type argument that the set of all such polynomials is indeed dense in $L^2$ (it's conveniently done with a functional version of the monotone class or $\pi$-$\lambda$ theorem). A key point is that because of the separability of $X$, one can show that $X^*$ generates the Borel $\sigma$-algebra of $X$. (Note that it is not necessary that $X^*$ be separable.)

This can be done in a more orderly way, too. We can find a countable sequence $e_i \in X^*$ which are $L^2$-orthonormal and span an $L^2$-dense subspace of $X^*$. Let $H_m(s)$ be the $m$th Hermite polynomial (e.g. $H_0(s) = 1$, $H_1(s) = s$, $H_2(s) = s^2 - 1$, etc) and consider the polynomials of the form
$$F_{a_1, \dots, a_k}(x) = H_{a_1}(e_1(x)) \cdots H_{a_k}(e_k(x)).$$
These functions form an orthogonal basis of $L^2$. Moreover, if we let $\mathcal{H}_n$ be the closed linear span of $\{F_{a_1, \dots, a_k} : a_1 + \dots + a_k = n\}$, then we get an orthogonal decomposition $L^2 = \bigoplus_{n=0}^\infty \mathcal{H}_n$. This is the so-called Wiener chaos decomposition. Note that $\mathcal{H}_n$ is naturally isomorphic to the $n$-fold symmetric tensor of $H$ with itself, and so we get an isomorphism of $L^2$ with the bosonic Fock space over $H$.

In particular, $\mathcal{H}_0 = C$ and $\mathcal{H}_1 = H$. So you were on the right track, but you stopped too soon :)

I wrote a little bit about this in these lecture notes, which also has a few references.