It is convenient to label irreps of the unitary group by highest-weight vectors $\lambda = (\lambda_1, \lambda_2,...\lambda_n)$ satisfying $\lambda_1 \geq \lambda_2\geq...\lambda_n)$. Then we obtain a formula for the characters (for diagonal elements) in terms of the Schur functions (with a small generalization allowing some $\lambda_j$ to be negative). When looking for characters of the projective unitary group it is natural to start with the characters of the unitary group which are invariant under phase shifts. By looking at the formula for characters we see that for phase-invariance what we need is $\sum_j \lambda_j = 0$ which is a very nice restriction.
My question is: is this "natural" guess correct - are the irreps of $\mathrm{PU}(n)$ those irreps of $\mathrm{U}(n)$ which happen to be invariant under phase shifts?
It appears the answer is yes, from relatively basic character theory. A representation $\rho$ of $\mathrm{PU}(n)$ gives you a representation $\tilde{\rho}$ of $\mathrm{U}(n)$ (just make it be constant on equivalence classes). Now compute
\begin{align} \int_{\mathrm{U}(n)} d\mu(g) \left\lvert\operatorname{tr}(\tilde{\rho}(g)\right\rvert^2 = \int_{\mathrm{PU}(n)} d\mu_P(x) \left\lvert\operatorname{tr}(\rho(x)\right\rvert^2 = 1. \end{align} Where we use the well-known fact that for phase invariant functions Haar integrals over the unitary group are the same as those over the projective unitary group. Now we use the fact that $\tilde{\rho}$ is irreducible iff the character is normalized in the way we just showed it is.