No, the dual operator $T'$ does not have a positive eigenvector, in general.
As a counterexample, consider the space $E = c_0(\mathbb{Z})$ if scalar-valued sequences indexed over the integers, endowed with the sup norm and the pointwise order. This is a Banach lattice and an M-space. Its dual space $E'$ can be identified with $\ell^1(\mathbb{Z})$.
Now let $L$ and $R$ denote the left and right shift on $E$, respectively, and define $T := (L+R)/2$.
Then $T$ is positive and irreducible and has norm $1$. The spectral radius of $T$ is $1$ (to see this, note that the bi-dual operator $T''$ on $\ell^\infty(\mathbb{Z})$ has the constant sequence with value $1$ as a fixed point).
However, $T' = (L' + R')/2$ does not have positive eigenvector. Indeed, assume to the contrary that $v \in E'$ is an eigenvector of $T'$ with entries $\ge 0$ and associated eigenvalue $\lambda$. Then $\lambda$ is a real number and $\lambda \ge 0$.
We distinguish between two cases:
case: $\lambda < 1$. In this case it follows that $T^nv \to 0$ as $n \to \infty$. Thus, the ideal
$$
I = \{x \in E': \; T^n |x| \to 0 \text{ as } n \to \infty\}
$$
in $E'$ contains the non-zero vector $v$. It is also closed (since $T'$ is power-bounded), and since the operator $T'$ can easily be checked to be irreducible, too, it follows that $I = E'$. But this cannot be true since $T'$ is norm preserving on the positive cone $E'_+$.
case: $\lambda = 1$.
In this case, we can use that $T'$ commutes with the right shift $L'$ and the left shift $R'$ on $E'$, and hence every shifted version of $v$ is also an eigenvector for the eigenvalue $1$. Therefore, the eigenspace of $T'$ for the eigenvalue $1$ is infinite-dimensional. However, it follows from infinite-dimensional Perron-Frobenius theory that the fixed space of a power-bounded irreducible operator is either $\{0\}$ or one-dimensional, so was also have a contradiction in this case.
