Let $A$ and $B$ be unital $C^*$-algebras. Let $a\in A$ and $b \in B$ be positive and invertible elements, so that in particular we can use the continuous functional calculus to define $a^z$ and $b^z$ and $(a\otimes b)^z$ where $z\in \mathbb{C}$. I want to show that $$(a\otimes b)^z = a^z\otimes b^z$$ as elements of the $C^*$-algebra $A\otimes B$ (say equiped with the minimal norm, but I don't think the choice of norm matters).
Attempt: Let $n \in \mathbb{N}\setminus \{0\}$. Then clearly $(a\otimes b)^{1/n} = a^{1/n} \otimes b^{1/n}$ by uniqueness of the positive $n^{th}$ root of a positive element. Hence, $(a\otimes b)^{m/n}= (a^{1/n}\otimes b^{1/n})^m = a^{m/n}\otimes b^{m/n}$ for all $m,n \in \mathbb{N}\setminus \{0\}$. Applying the same argument with $a$ replaced by $a^{-1}$ and $b$ by $b^{-1}$ yields $(a\otimes b)^q = a^q \otimes b^q$ for all $q\in \mathbb{Q}$. By density of $\mathbb{Q}$ in $\mathbb{R}$ and using the fact that $x \mapsto x^z$ is uniformly continuous on compacts, I can then deduce that $(a\otimes b)^r = a^r \otimes b^r$ for all $r\in \mathbb{R}$.
How can I deduce that the equality holds for all complex numbers? Perhaps we must prove that $$z \mapsto (a\otimes b)^z, \quad z \mapsto a^z \otimes b^z$$ are analytic functions $\mathbb{C}\to A\otimes B$ that agree on $\mathbb{R}$, so that they agree everywhere?
The argument with the analytic functions probably works, but I'm usually scared of operator-valued analytic functions, so I will not go there.
On the other hand, you can do this kind of directly.
First, since $(a\otimes 1)^n=a^n\otimes 1$, we get that $p(a\otimes 1)=p(a)\otimes 1$ for any polynomial, and this forces $$\tag1 f(a\otimes 1)=f(a)\otimes 1,\qquad f(1\otimes b)=1\otimes f(b) $$ for any continuous function $f$.
Now, by definition, \begin{align}\tag2 (a\otimes b)^z=e^{z\,\log(a\otimes b)}. \end{align} Note that, since $a\otimes 1$ and $1\otimes b $ commute, their logarithms commute and $$\tag3 e^{\log (a\otimes 1) + \log(1\otimes b)}=e^{\log (a\otimes 1)}e^{\log(1\otimes b)}=(a\otimes 1)(1\otimes b)=a\otimes b. $$ Thus, as the exponential is injective on positive operators, \begin{align}\tag4 \log (a\otimes 1)+\log (1\otimes b)=\log (a\otimes b). \end{align} Combining everything, \begin{align} (a\otimes b)^z&=e^{z\log(a\otimes b)}=e^{z(\log a \otimes 1)+z(1\otimes\log b)}=e^{z\log(a\otimes 1)}\,e^{z\log(1\otimes b)}\\[0.3cm] &=(a\otimes 1)^z(1\otimes b)^z =(a^z\otimes 1)(1\otimes b^z)=a^z\otimes b^z \end{align}