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.
Best Answer
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.