$(a\otimes b)^z = a^z \otimes b^z$ for elements in a $C^*$-algebra and $z\in \mathbb{C}$.

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}