If you want to understand $S^{-1}A$ for any $S\subset A$, you may write $S$ as a filtered union of its finite subsets $S_i$, and it is clear from the universal properties that
$$S^{-1}A=\varinjlim_i \ S_i^{-1}A$$
Therefore it is sufficient to consider the case where $S$ consists of a finite set of elements of $A$. If $f$ denotes the product of all the elements of $S$ in $A$, it is clear that inverting $f$ is the same thing than inverting each element of $S$. Therefore, we may assume that $S$ contains a unique element $f$. It is then obvious (in terms of universal properties) that $S^{-1}A$ is canonically isomorphic to the (filtered) colimit of the diagram indexed by non negative integers
$$A\overset{f}{\to}A\overset{f}{\to}A\overset{f}{\to}\cdots\overset{f}{\to}A\overset{f}{\to}A\overset{f}{\to}\cdots$$
(where $f$ stands for `multiplication by $f$').
Puting back all of the above reductions/descriptions together is the categorical way of writing the usual description of $S^{-1}A$ with elements. (As an exercise, you may turn all this into a global construction, i.e. as one filtered colimit of a diagram which is objectwise just $A$, but in which the transition maps are given by multiplication by a finite product of elements of $S$.)
The problem is now reduced to the understanding of colimits of diagrams of $A$-modules
of the shape
$$M_0\overset{s_1}{\to} M_1 \overset{s_2}{\to} M_2\to\cdots
M_n\overset{s_{n+1}}{\to} M_{n+1}\to\cdots$$
Using the universal properties, one can see that $\varinjlim_n M_n$
is the cokernel of the map
$$1-s:\bigoplus_n M_n\to \bigoplus_n M_n$$
where $s$ sends an element $x$ of $M_n$ to $s_{n+1}(x)$.
In particular, we have a canonical epimorphism
$$\bigoplus_n M_n\to \varinjlim_n M_n$$
This implies that any element of $\varinjlim_n M_n$ comes from an element of $M_n$
for $n$ big enough. This presentation of $\varinjlim_n M_n$
also shows that if an element $x$ of $M_n$ becomes zero in $\varinjlim_n M_n$,
there exists a positive integer $m$ such that $s_{n+m}\ldots s_{n+1}(x)=0$.
Applying this to the diagram
$$A\overset{f}{\to}A\overset{f}{\to}A\overset{f}{\to}\cdots\overset{f}{\to}A\overset{f}{\to}A\overset{f}{\to}\cdots$$
we see that $A[f^{-1}]$ admits the usual description.
N.B. In an abstract context, proving things about localizations consists to use the categorical description above and to use some special properties of filtered colimits (which are often exact, for instance), so that you don't need any description in terms of elements. For instance, the flatness of $S^{-1}A$ comes from the fact that it is a filtered colimit of free $A$-modules (of rank $1$).
One can also compute the kernel of the map $\tau:A\to S^{-1}A$. To keep things simple let us do this in the case where $S$ consists of a single element $f$. Then $ker(\tau)$ is the colimit of $A$-modules
$$\varinjlim_n \; \; ker(A\overset{f^n}{\to}A)$$
(because filtered colimits are exact in the category of $A$-modules).
In conclusion, it seems possible to describe $S^{-1}A$ using categorical arguments for $A$-modules, but I don't see how to obtain such a description using only the theory of rings.
This question was posted some time ago but it's never been answered properly, so it seems worthwhile to record the answer anyway. In short, when $A$ is finitely generated, the associated graded is generated by the elements you describe if and only if the filtration is inherited from the natural grading of a free associative algebra.
Suppose that $A$ is filtered $A = \bigcup_{i\geq 0} A_i$ with $A_0 = k$ and that $A$ is generated by a finite set of elements, say $x_1,...,x_n$. We may suppose that $x_i \in A_{d_i}\setminus A_{d_i-1}.$ By the universal property of free associative algebras there is a homomorphism from $F = k\langle X_1,...,X_n\rangle$ onto $A$, say $\phi: F \twoheadrightarrow A$ which sends $X_i \mapsto x_i$. Now $F$ is also graded by placing $X_i$ in degree $d_i$ and this defines a filtration $F = \bigcup_{k \geq 0} F_k$ where $F_k$ is the span of monomials $x_{i_1} \cdots x_{i_m}$ with $\sum_{j=1}^m d_{i_j} \leq k$ and $1\leq i_1,...,i_m \leq n$. Note that $\phi$ preserves the filtration.
The image of $\operatorname{gr}\phi$ is the subalgebra of $\operatorname{gr} A$ generated by the elements $x_i + A_{d_i-1}$ and so your question can now be phrased as asking when is $\operatorname{gr} \phi$ surjective? According to Corollary 7.6.14 of McConnell and Robson "Noncommutative Noetherian rings" this is the case if and only if $A_i = \phi(F_i)$. This also tells us that, out of all the filtrations on $A$ satisfying $x_i \in A_{d_i}\setminus A_{d_i-1}$ there is a unique filtration such that $\operatorname{gr} A$ is generated by the elements $x_i + A_{d_i-1}$, and it is the minimal filtration in a precise sense.
Now consider the example of the three dimensional Heisenberg algebra $\mathfrak{h}$ which you mentioned. We can view this as a quotient of the free algebra $F = k\langle X, Y\rangle$ by the ideal generated by $[X,[X,Y]]$ and $[Y, [X, Y]]$, however if you want the element $z := \phi [X, Y]$ to lie in degree 1 of $U(\mathfrak{h})$ (in accordance with the PBW theorem) then you have $\phi(F_1)$ spanned by ${1, \phi(X), \phi(Y)}$, whilst in $U(\mathfrak{h})$ the degree 1 filtered component also contains $z$.
Best Answer
The associated graded of a filtered R-module M is the universal R-module with a map of the Rees module of M over R[t] to gr M.
Let me explain what the Rees module Rees(M) is: it's the submodule of M[t,t-1] which is generated as a R[t] module by tiM_i. Give this the obvious grading by degree of t. So Rees(M)/tRees(M)=gr M, whereas Rees(M)/(t-1)Rees(M)=M with the induced filtration. This is the thing that has a map to gr M.