Given two cohomology theories $h^{\bullet}$ and $k^{\bullet}$, we define their direct sum $(h \oplus k)^{\bullet}$ as the cohomology theory $(h \oplus k)^{n}(X) := h^{n}(X) \oplus k^{n}(X)$. If I'm not wrong, this is another cohomology theory, whose spectrum is the wedge product of the corresponding spectra. For example, the singular cohomology with coefficients in $\mathbb{Z} \oplus \mathbb{Z}$ is the direct sum of two copies of the one with coefficients in $\mathbb{Z}$.

Let us call a theory $h^{\bullet}$ irreducible if it is not decomposable as a non-trivial direct sum. Is it true that, if $h^{\bullet}$ is irreducible, then the cohomology groups of the point are also irreducible? In other words, is it true that $h^{n}(pt) = \mathbb{Z}, \mathbb{Z}/p\mathbb{Z}$ or $0$, but never a non-trivial direct sum of them?

If the answer is no, is it true for multiplicative theories?

Added later: The answer is no, as shown by Eric Wofsey. So I modify the question: is it true that the free part of $h^{n}(pt)$ is $\mathbb{Z}$ or $0$, but never $\mathbb{Z}^{n}$ with $n > 1$?