I think the "modified conjecture" (that is, one gets a homomorphism or anti-homomorphism) can be proved by elementary calculations, in this setting. (Walter's theorem, which Yemon alludes to, uses only that we have an isometry, not that the map has the rather special form). However, my tolerance for such calculations is low at the moment (and I cannot see how to organise the data in a "clever" way) so I'll just indicate the technique and some partial results.

The idea is that $a\in\mathbb CH$ is positive if and only if $(a\xi|\xi)\geq 0$ for all $\xi\in\ell^2(H)$. We aim to choose $\xi$ intelligently, but first we make a general observation. Let $(\delta_h)_{h\in H}$ be the standard orthonormal basis of $\ell^2(H)$ and let $\lambda_x$ be the left-multiplication operator given by $x\in H$. Then
$$ (\lambda_x\xi|\xi) = \sum_{h\in H} \xi_{x^{-1}h} \overline{\xi_h} =
\sum_h \overline{\xi_h} \check\xi_{h^{-1}x} = \overline\xi \star \check\xi = \varphi(x), $$
say. I call such $\varphi$ *positive definite*. Here $\check\xi_h = \xi_{h^{-1}}$ for each $h$, and $\star$ is convolution.

Choose $\xi = \delta_e + u\delta_h$ for $u\in\mathbb C$, so $\check\xi = \delta_e + u\delta_{h^{-1}}$ and thus
$$ \varphi(x) = (\overline{\xi} \star \check\xi)(x)
= (1 + |u|^2)\delta_{x,e} + u\delta_{x,h^{-1}} + \overline{u} \delta_{x,h}. $$

Claim: $f(x^{-1}) = f(x)^{-1}$ for each $x\in G$.

I write $1$ for $\lambda_e$.
To show this, consider $a = (1+z\lambda_x)^\ast (1+z\lambda_x) \geq 0$ in $\mathbb CG$, where $z\in\mathbb C$. Then
$$ f(a) = (1+|z|^2) + z\lambda_{f(x)} + \overline{z} \lambda_{f(x^{-1})}. $$
By hypothesis, $f(a)\geq 0$ so $\langle f(a), \varphi \rangle \geq 0$. Towards a contradiction, suppose $f(x^{-1})\not=f(x)^{-1}$.

- If $f(x)=e$ then choose $h=e$ so $\varphi(x) = t\delta_{x,e}$ for some $t\geq 0$, so
$$ \langle f(a), \varphi \rangle = t(1+|z|^2+z) $$
which is not positive for all $z\in\mathbb C$.
- If $f(x)\not=e$ but $f(x^{-1})=e$ then argue similarly.
- So $f(x)\not=e$ and $f(x^{-1})\not=e$. Set $h=f(x)$ so then
$$ \langle f(a), \varphi \rangle = (1+|u|^2)(1+|z|^2) + \overline{u}z
+\overline{uz}\delta_{f(x), f(x^{-1})} + uz\delta_{f(x)^{-1},f(x)}. $$
- Regardless of the values of $\delta_{f(x), f(x^{-1})}$ and $\delta_{f(x)^{-1},f(x)}$, this can never be positive for all $z,u$.
- So the claim holds.