Last time I checked, the entire function $f \colon \mathbf C \to \mathbf C \colon z \mapsto a\exp(iz)$ was transcendental for every non-zero $a \in \mathbf C$. So, taking $\alpha = -a = 2\pi$ yields a negative answer for the "$\alpha + f(\alpha)$" case; taking $\alpha = a^{-1} = 2\pi$ yields a negative answer for the "$\alpha f(\alpha)$" case; and taking $\alpha = a = 2\pi$ yields a negative answer for the "$\alpha/f(\alpha)$" case. The "$\alpha - f(\alpha)$" case is a trivial variant of the "$\alpha + f(\alpha)$" case.
Edit. To address the question in the first comment under this answer, the function $f \colon \mathbf R \to \mathbf R \colon x \mapsto a \cos x$ is transcendental for every non-zero $a \in \bf R$, with $f(2\pi) = a$. So again, taking $\alpha = -a = 2\pi$ yields a negative answer for the "$\alpha + f(\alpha)$" case; taking $\alpha = a^{-1} = 2\pi$ yields a negative answer for the "$\alpha f(\alpha)$" case; and taking $\alpha = a = 2\pi$ yields a negative answer for the "$\alpha/f(\alpha)$" case.