It is dimensionless in the sense of mass dimension.
Setting $\hbar = c = 1$ means we only need to fix one base unit, which is usually taken to be the energy measured in $\mathrm{eV}$. Now, since $c=1$, this means that because of $E=mc^2$ becoming $E=m$ both $E$ and $m$ are measured in $\mathrm{eV}$. They might represent different dimensions (mass and energy), but they are measured in the same unit. Now, $E=h\nu$ means that inverse time is measured in $\mathrm{eV}$, so time is measured in $\mathrm{eV^{-1}}$. And so on.
Now, the "mass dimension" of a quantity is simply the power of $\mathrm{eV}$ it is measured in. Since the action is the integral of an energy against time, it has units of $\mathrm{eV}\cdot\mathrm{eV}^{-1} = \mathrm{eV}^0$, i.e. it has mass dimension zero.
You are right in that it is not "dimensionless". But having mass dimension zero means for any quantity $Q$ that there are powers of $\hbar$ and $c$ such that $\frac{Q}{\hbar^n c^m}$ is dimensionless, and since $\hbar = c = 1$, $\frac{Q}{\hbar^n c^m} = Q$, so there is no numerical difference between those quantities, and one sloppily says that $Q$ is dimensionless.
If you are somewhat worried that $\frac{Q}{\hbar^n c^m} = Q$ "looks wrong" from a dimensional analysis standpoint, then yes, that's right - the convenience in the formulae we get from $\hbar = c = 1$ comes inevitably with the loss of a large part of dimensional analysis, all that is left is the mass dimension for that.