Question: Suppose we have a function $f: A\to B$. Then we know, without specifying what $f$ is, that $f$ may or may not map to every element $b\in B$. If $f$ does map to every element $b\in B$ then it's surjective. 

However, does the notation $f:A\to B$ imply that every element $a\in A$ is mapped to some unique $b\in B$? 

Furthermore, is there a difference between domain and natural domain? If there is, how can we tell whether $A$ is a/the natural domain from the notation $f: A\to B$?

Suppose we consider the function defined this way: $f: \mathbb{R}\to \mathbb{R}, f=\frac{1}{x}$ Clearly, $f$ is not defined at $x=0$. Would it therefore be incorrect to write $f: \mathbb{R}\to \mathbb{R}, f=\frac{1}{x}$ instead of $f: \mathbb{R}\backslash \{0\}\to \mathbb{R}, f=\frac{1}{x}$?

*The question is a bit clumsy. You are welcome to ask for clarification if this isn't clear enough.*

Answer: 
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> Suppose we have a function $f: A\to B$. Then we know, without specifying what $f$ is, that $f$ may or may not map to every element $b\in B$. If $f$ does map to every element $b\in B$ then it's surjective. 
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That is correct.

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> However, does the notation $f:A\to B$ imply that every element $a\in A$ is mapped to some unique $b\in B$? 
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Depends on what you mean by unique. Each $a \in A$ maps to exactly 1 $b \in B$. But each $b \in B$ may be mapped from any number of $a \in A$, or not be mapped from any at all.

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> Furthermore, is there a difference between domain and natural domain? 
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Natural domain is the largest set of values for which a function is defined, which, ironically, isn't a well defined concept. 

Sometimes natural domain will be used to mean "all real numbers for which the function is defined" or "all complex numbers for which a function is defined".

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> If there is, how can we tell whether $A$ is a/the natural domain from the notation $f: A\to B$? 
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It depends on which definition of "natural domain" you are using. If you can determine that the natural domain is equal to $A$, then the natural domain is equal to $A$. If you can determine that it is not equal, then it is not equal.

For example, under the definition of natural domain being "all real numbers for which a function is defined", you then 

$$f(x) = x^{-1} ~:~ \mathbb R^{+} \to \mathbb R$$

Here $A=\mathbb R^{+}$ is not the natural domain, since the function could have been defined for $R^{-}$ as well. The natural domain is not $\mathbb R$, since the function isn't defined for $0$.
In general, the notation $f:A \to B$ implies that every element $a \in A$ is mapped to **some** element $b∈B$, which is not necessarily unique. 

However, the definition of a function means that this element *is* unique *in terms of a*, ie. for any $a\in A$ there exists only **one** $b \in B$ such that $f(a)=b$. If you are requiring uniqueness *in terms of b* as well, ie. for any $b \in B$ there can exist only **one** $a \in A$ such that $f(a)=b$, then the mapping is said to be injective. To write this formally, you could state:

\begin{align}
f(x)=f(x') \Rightarrow x = x'
\end{align}

This does not necessarily mean that every element of $B$ **does** have a corresponing element from $A$; It is very easily to see that the domain {1,2} can map injectively to {1,2,3} with each number in $A$ mapping to itself in $B$. This would meet the criteria for being an injective function, but would not be surjective.

Regarding domain vs. natural domain, I believe that the *natural* *domain* is the largest domain for which the mapping is defined. Every other domain must necessarily be a subset of the *natural domain*. 

In your example, $A$ would be the *natural domain* as $f$ maps every single element in $A$ to a corresponding element in $B$, and any other domain will undoubtedly be a subset of it - the function is not specified to map elements outside of $A$ because we don't know what $A$ actually is. 

As a counter-example, if we knew that $f: \mathbb R \to \mathbb R$, and were *then* given $f: A\to B$ with $A \subset \mathbb R$, then A would **not** be the *natural domain* any longer, as we would be aware that there exists a larger domain for which the function operates, ie. $\mathbb R$.