Can I "naïvely" claim that (complex) Taylor series are all analytic because each one of their terms is clearly entire?
The reason for my hesitation is because there are infinitely many terms involved, so it's not clear to me why such a justification must necessarily hold.
Sure, we're handling series over their convergence discs, but why does that matter?
You have your definitions wrong, and the comments seem to suggest this as well.
A function $f$ defined on an open subset $\mathrm{O}$ of the normed vector space $\mathbf{F}^p$ (where $\mathbf{F}$ is either the real or complex field) with values in any normed vector space $\mathrm{X}$ (over the same field $\mathbf{F}$) is analytic if, for whatever point $o \in \mathrm{O},$ there exists a neighbourhood of $o$ and a family of vectors $(c_\nu)_{\nu \in \mathbf{Z}_+^p}$ such that the family $(c_\nu (z-o)^\nu)$ is absolutely summable and its sum coincides with $f(z)$ ($z$ belonging to the neighbourhood mentioned), where for any vector $t = (t_1, \ldots, t_p) \in \mathbf{F}^p$ we write $t^\nu = t_1^{\nu_1} \cdots t_p^{\nu_p} \in \mathbf{F},$ a scalar.
In the particular case in which $\mathbf{F} = \mathbf{C}$ and $\mathrm{X} = \mathbf{C}^q$ we have the celebrated theorem of Hartog-Goursat-Cauchy which says that a function $f$ is analytic on an open set if and only if it has partial derivatives.
To boil it down to $p = q = 1,$ we have that a complex-valued function defined on an open subset of the complex plane is analytic if and only if it is differentiable (or as is sometimes say: "holomorphic").
To relate this to your question, if you give any power series $(a_n z^n)_{n \in \mathbf{Z}_+}$ in one complex variable, then it will define an analytic function wherever the series converges. If the series converges on an open set (which can be proven to be necessarily some disc) around zero and $f$ is the function so defined, then the Taylor's coefficients of $f$ are $a_n$ because $f^{(n)}(0)=n!a_n$ from basic differentiation of absolutely sure power series.