To expand a bit on my comment above:
Being isomorphic as a locally ringed space to $(\mathbb{R}^n,\mathcal{O})$ doesn't impose additional conditions on the underlying topological space of a locally ringed space beyond requiring it to be locally homeomorphic to $\mathbb{R}^n$. (Well, that's a lie: a differentiable structure does of course place limitations on the topology of a manifold, but only very subtle ones: it doesn't impose either of the limitations you are asking about. See below!)
Thus, if you want your definition of a manifold to include Hausdorff and second countable and/or paracompact, you had better put that in explicitly. (And, although it's a matter of taste and terminology, in my opinion you do want this.)
I think you will find these lecture notes enlightening on these points. In particular, on page 4 I give an example (taken from Thurston's book on 3-manifolds!) of a Galois covering map where the total space is a manifold but the quotient space is not Hausdorff. (When I gave this example I mentioned that I wish someone had told me that covering maps could destroy the Hausdorff property! And indeed the audience looked suitably shaken.)
With regard to your other question ("Also, is this definition given carefully in any textbook?")...I completely sympathize. When I was giving these lectures I found that I really wanted to speak in terms of locally ringed spaces! See in particular Theorem 9 in my notes, which contains the unpleasantly anemic statement: "If $X$ has extra local structure, then $\Gamma \backslash X$ canonically
inherits this structure." What I really wanted to say is that if $\pi: X \rightarrow \Gamma \backslash X$, then $\mathcal{O}_{\Gamma \backslash X} = \pi_* \mathcal{O}_X$! (I am actually not the kind of arithmetic geometer who has to express everything in sheaf-theoretic language, but come on -- this is clearly the way to go in this instance: that one little equation is worth a thousand words and a lot of hand waving about "local structure".)
What is even more ironic is that my course is being taken by students almost all of whom have taken a full course on sheaves in the context of algebraic geometry. But whatever differential / complex geometry / topology they know, they know in the classical language of coordinate charts and matrices of partial derivatives. It's really kind of a strange situation.
I fantasize about teaching a year long graduate course called "modern geometry" where we start off with locally ringed spaces and use them in the topological / smooth / complex analytic / Riemannian categories as well as just for technical, foundational things in a third course in algebraic geometry. (As for most graduate courses I want to teach, improving my own understanding is a not-so-secret ulterior motive.) In recent years many similar fantasies have come true, but this time there are two additional hurdles: (i) this course cuts transversally across several disciplines so implicitly "competes" with other graduate courses we offer and (ii) this should be a course for early career students, and at a less than completely fancy place like UGA such a highbrow approach would, um, raise many eyebrows.
Yes: Let $(f,\psi):X\to Y$ be a morphism of locally ringed spaces, where $X$ and $Y$ are smooth manifolds with their sheaves of smooth functions. If $\psi:C^\infty_Y \to f_* C^\infty_X$ is a morphism of sheaves of $\mathbb R$-algebras, then $f$ is smooth and $\psi=f^\#$.
Proof. Let $s:U\to \mathbb R$ be a smooth function. The equation $\psi s= s\circ f$ follows from the commutativity of the diagram below. Notice the triangle commutes because there is a unique $\mathbb R$-algebra map $C^\infty_{f(x)}/{\frak m}_{f(x)}\cong \mathbb R \to \mathbb R$. It now follows that $f:X\to Y$ is smooth. Indeed, we know $s\circ f$ is smooth for all real valued functions $s$ on $Y$, and we may take $s$ to be the coordinate functions of charts on $Y$. QED.
Best Answer
Isomorphisms of locally ringed spaces are the same as isomorphisms of ringed spaces (which happen to be locally ringed), so you do not have to specify that $(M,\mathcal{O}_M)$ is locally ringed. That is, if $(M,\mathcal{O}_M)$ is a ringed space which is locally isomorphic to $(\mathbb{R}^n, \mathcal{O})$, then the stalks of $\mathcal{O}_M$ are isomorphic to the corresponding stalks of $\mathcal{O}$, and thus are local rings. The local isomorphisms from $(M,\mathcal{O}_M)$ to $(\mathbb{R}^n, \mathcal{O})$ are then automatically isomorphisms of locally ringed spaces (basically, because the condition for a morphism to be "local" involves only the ringed space structure, and so is satisfied by any isomorphism of ringed spaces).
So you can define a manifold as a ringed space that is locally isomorphic to $(\mathbb{R}^n, \mathcal{O})$. However, it is better to define it as a locally ringed space because smooth maps between manifolds correspond to morphisms of locally ringed spaces, not morphisms of ringed spaces. I don't know a counterexample off the top of my head (it seems rather complicated to construct one), but there isn't any reason to expect that a morphism of ringed spaces between two manifolds is the same thing as a smooth map. The local condition on morphisms says that pullback respects evaluation of smooth functions at points: that is, if $\varphi:(M,\mathcal{O}_M)\to(N,\mathcal{O}_N)$ is a morphism, $f\in\mathcal{O}_N(U)$, and $p\in \varphi^{-1}(U)$, then $(\varphi^*f)(p)=f(\varphi(p))$. In this way, the pullback map $\varphi^*$ on the sheaves is completely determined by the map of sets $M\to N$. If you drop this condition (i.e., you talk only about morphisms of ringed spaces instead of morphisms of locally ringed spaces), it seems very difficult to control what the pullback map $\varphi^*$ can possibly look like.
So to sum up: you don't need to say "locally ringed" instead of "ringed" when talking about the objects of the category of manifolds, but you (probably) do need to say it when talking about the morphisms (though I don't know a counterexample that would prove this).