Yes. Let $T$ denote the transition kernel. Suppose $\pi$ and $\nu$ are distinct stationary distributions. Let $\tau'$ be defined by $\tau'(A) = \min(\pi(A),\nu(A))$: then $\tau'$ is a finite measure. Let $p = \tau'(\mathbb{R}^+)$; since $\pi \neq \nu$, we have $p < 1$, and since the chain is irreducible, $p > 0$. Let $\tau = \tau'/p$, $\tilde{\pi} = (\pi-\tau')/(1-p)$ and $\tilde{\nu} = (\nu - \tau')/(1-p)$, so that

$\pi = p\tau + (1-p)\tilde{\pi}$

and

$\nu = p\tau + (1-p)\tilde{\nu}$.

Note that $\tilde{\pi} = \max(0, \pi-\nu)/(1-p)$ and $\tilde{\nu} = \max(0,\nu-\pi)/(1-p)$, so that $\max(\tilde{\pi},\tilde{\nu}) = 0$

Now we show that $\tau$ is also a stationary distribution of the chain. Observe that $\pi= T\pi = pT\tau + (1-p)T\tilde{\pi}$ so $\pi \geq pT\tau$. Similarly, $\nu \geq pT\tau$. Therefore, $pT\tau \leq \min(\pi,\nu) = p\tau$. Combined with the fact that $\tau(\mathbb{R}^+) = T\tau(\mathbb{R}^+)$, we have $T\tau = \tau$.

This, in turn, implies that $\tilde{\pi}$ and $\tilde{\nu}$ are also stationary distributions. However, the fact that $\max(\tilde{\pi},\tilde{\nu}) = 0$ then implies that the chain is reducible--a contradiction.