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Algebraically, is taking limits an operator? If so, unlike many operators, it satisfies many operations like $$\lim_{\dots}(\log(x))=\log(\lim_{\dots} x).$$ I mean, not all operations satisfy this property, so is there an algebraic classification of such operators?

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In some sense it is an operator. Let $X$ be a topological space and $p \in X$. Let $R$ be the set of all functions $f : X \to \mathbb{R}$ such that $\lim_{x \to p} f(x)$ exists. Actually this is a subring of the ring of all functions, and $f \mapsto \lim_{x \to p} f(x)$ is a ring homomorphism $R \to \mathbb{R}$. For functions $f$ which are continuous at $p$, this coincides with the evaluation homomorphism $f \mapsto f(p)$.

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The limit is a family of linear functionals on the space of functions which are continuous everywhere (so the limit is well-defined).

Every $x\in\Bbb R$ (or another space) defines a functional $\lim\limits_{t\to x} f(t)$. But because $f$ is continuous this is really just $f(\lim_{t\to x} t)=f(x)$. So those are the usual evaluation functionals.

The idea is that the continuity of the functions assures that this is really a well-defined operator, and that its result is the evaluation functional at the point.

You can talk about functions with certain points of discontinuities, or one-sided limits, but in all of them some weak form of continuity must play a role, or else the limit is not well-defined and it won't be a functional (which is a sort of operator).

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Think about continuity; this is the essential property here.

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Algebraically, is limit a operator?

What is an operator in the context of this question?

The question you are asking is : "is limit of a function is same as function of a limit"?

The answer is no, "limit of a function is not always same as function of a limit"

Learn the $\epsilon,\delta $ definition of limit

The name for this property is commutative :

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