I would like to give some details in order to make clear that one can give a proof with hardly any computations at all (I have never looked at the Bourbaki presentation but I guess they make the same point though, because they want to make all proofs only depend on previous material they might make a few more computations).
To begin with we are considering supercommutative algebras (over a commutative base ring $R$) which are strictly commutative, i.e., $\mathbb Z/2$-graded algebras with $xy=(-1)^{mn}yx$, where $m$ and $n$ are the degrees of $x$ and $y$ and $x^2=0$ if $x$ has odd degree. Note that the base extension of such an algebra is of the same type (the most computational part of such a verification is for the strictness which uses that on odd elements $x^2$ is a quadratic form with trivial associated bilinear form). The exterior algebra is then the free strictly supercommutative algebra on the module $V$. Furthermore, the (graded) tensor product of two sca's is an sca and from that it follows formally that $\Lambda^\ast(U\bigoplus V)=\Lambda^\ast U\bigotimes\Lambda^\ast V$.
Now, the diagonal map $U\to U\bigoplus U$ then induces a coproduct on $\Lambda^\ast U$ which by functoriality is cocommutative making the exterior algebra into superbialgebra. I also want to make the remark that if $U$ is f.g. projective then so is $\Lambda^\ast U $. Indeed, by presenting $U$ as a direct factor ina free f.g. module one reduces to the case when $U$ is free and then by the fact that direct sums are taken to tensor products to the case when $U=R$ but in that case it is clear that $\Lambda^\ast U=R\bigoplus R\delta$ where $\delta$ is an
odd element of square $0$.
If now $U$ is still f.g. projective then $(\Lambda^\ast U^\ast)^\ast$ is also a supercommutative and supercocommutative superbialgebra. We need to know that it is strictly supercommutative. This can be done by a calculation but we want to get by with as much handwaving as possible. We can again reduce to the case when $U$ is free. After that we can reduce to the case when $U=R$ (using that the tensor product of sca's is an sca) and there it is clear (and one can even in that case avoid computations). Another way, once we are then free case, is to use the base change property to reduce to the when $R=\mathbb Z$ and then we have seen that the exterior algebra is torsion free and a torsion free supercommutative algebra is an sca.
We also need to know that if $U$ is f.g. projective, then the structure map $U\to\Lambda^\ast U$ has a canonical splitting. This is most easily seen by introducing a $\mathbb Z$-grading extending the $\mathbb Z/2$-grading and showing that the degree $1$ part of it is exactly $U$ (maybe better would have to work from the start with $\mathbb Z$-gradings).
This splitting gives us a map $U\to(\Lambda^\ast U^\ast)^\ast$ into the odd part and as we have an sca we get a map $\Lambda^\ast U\to(\Lambda^\ast U^\ast)^\ast$ of sca's. This map is an isomorphism. Indeed, we first reduce to the case when $R$ is free (this is a little bit tricky if one wants to present $U$ as a direct factor of free but one may also use base change compatibility to reduce to the case when $R$ is local in which case $U$ is always free). Then as before one reduces to $U=R$ where itt is clear as both sides are concentrated in degrees $0$ and $1$ and the map clearly is an iso in degree $0$ and by construction in degree $1$.
Note that everything works exactly the same if one works with ordinary commutativity (so that the exterior algebra is replaced by the symmetric one) up till the verification in the case when $U=R$. In that case the dual algebra is the divided power algebra and we only have an isomorphism in characteristic $0$.
Afterthought: The fact that things work for the exterior algebra but introduces divided powers in the symmetric case has an intriguing twist. One can define divided power superalgebras where all higher powers of odd elements are defined to be zero. With that definition the exterior algebra has a canonical divided power structure (for a projective module) which is characterised by being natural and commuting with base change (or being compatible with tensor products, just as for ordinary divided powers there is a natural divided power on the tensor product of two divided power algebras). Hence, somehow the fact that the exterior algebra is selfdual is connected with the fact that the exterior algebra is also the free divided power superalgebra on odd elements. However, I do not know of any a priori reason why the dual algebra, both in the symmetric and exterior case, should have a divided power structure.
As a curious aside, the divided power structure on the exterior algebra is related to the Riemann-Roch formula for abelian varieties: If $L$ is a line bundle on an abelian variety the RR formula says that
$$
\chi(L) = \frac{L^g}{g!}
$$
and this gives (as usual for Aityah-Singer index formulas) also the integrality statement that $L^g$ is divisible by $g!$. However, that follows directly from the fact that the cohomology of an abelian variety is the exterior algebra on $H^1$ and the fact that the exterior algebra has a divided power structure.
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
Too long for a comment, I was wondering about the cost of completely unfolding Darij's excellent argument. In fact, we have a characterisation of that sort of perturbations of identity that are isomorphisms. The statement is as follows
Lemma. Let $k$ be a field and $x$ an indeterminate. For every $Q\in k[x]$, let $f_Q$ be the morphism of $k[x]$ sending $x$ to $x+x^2Q$. Then $f_Q$ is an isomorphism iff $Q=0$.
Proof. One way being obvious (if $Q=0$ $f_Q=Id$), it is sufficient to prove that, if $Q\not=0$ then $x\notin Im(f_Q)=f_Q(k[x])$.
It is not difficult to see that $Im(f_Q)$ is a subalgebra as follows $$Im(f_Q)\subset k\oplus (x+x^2Q)k[x]$$ which does not contain $x$ (indeed $P\in Im(f_Q)$ implies that $P(x)-P(0)$ can be divided by $(x+x^2Q)$ which is true for $P=x$ iff $Q=0$).