Suppose that $H\mapsto U$ and $Hv=\lambda v$. Then for 1) $Uv = \frac{\lambda+i}{\lambda-i}v$ and for 2) $Uv = \exp(i\lambda)v$. Let $Uv=\kappa v$.
For 1) as $\lambda \rightarrow \infty$, $\kappa \rightarrow 1$. However, $\kappa$ never attains the value of $1$ and so the Cayley Transform does not map Hermitian matrices onto the unitary matrices (it is, however, one-to-one).
For 2) $\exp(i(\lambda + 2n\pi))=exp(i\lambda))$. So the matrix exponential is not one-to-one (it is, however, onto).
In the comments it is mentioned that a bijection is desired. We limit ourselves to a class of functions that preserve eigenvectors while mapping real eigenvalues to complex ones of modulus 1. This would include the power series in $H$, as assumed in Nick's answer. Thus, we can simply look for bijective maps from $\mathbb{R}\rightarrow S^1$.
Since $\mathbb{R}$ and $S^1$ are not homeomorphic, there are no bi-continuous bijections between the two.
There are, however, discontinuous bijections between the two sets (see this question for bijection from $(0,1) \rightarrow (0,1]$, which gives us a bijection $\mathbb{R}\rightarrow S^1$. However, in this example, there are countably infinite discontinuities, so it is not a particularly nice bijection).
Without rigor, I would say that none of the power series mappings would be invertible on the basis that the series would need to converge everywhere (or the mapping is ill-defined) and therefore will be continuous everywhere. Since our mapping is defined by the power series, it is analytic, hence we can find a power series for the inverse function by the Lagrange inversion theorem (there is possibly an issue if the derivative at a point is 0, but since the function is invertible it hopefully isn't a problem). Therefore, the inverse is continuous. Therefore, if a power series is a bijection it is bicontinous. This is a contradiction (as $\mathbb{R}$ and $S^1$ are not homeomorphic). Therefore, no power series in $H$ gives a bijection from the Hermitian to unitary matrices.