It's not very hard to just do this explicitly, along the lines you started but keeping my comment in mind. Rather than $h, e, f$ I'll write $H, X, Y$. Suppose $I$ is a nonempty ideal. Then it contains some nonzero element $aH + bX + cY$. Applying $[H, -]$ to this element gives
$$b [H, X] + c [H, Y] = 2b X - 2c Y$$
from which we conclude that either $a = 0$ or $I$ contains $H$. If $I$ contains $H$, then it contains $[H, X] = 2X$ and $[H, Y] = 2Y$, and hence must be all of $\mathfrak{g} = \mathfrak{sl}_2(\mathbb{C})$.
Otherwise, $a = 0$. Now, applying $[X, -]$ to $bX - cY$ gives
$$-c [X, Y] = -c H$$
from which we conclude that either $c = 0$ or $I$ contains $H$. If $I$ contains $H$ then as above we are done; otherwise, $I$ contains $bX$, so it contains $[X, Y] = H$, and again we're done.
A slicker way to organize this argument is as follows. Any ideal of $\mathfrak{g}$ must in particular be invariant under the adjoint action of $H$. The adjoint action of $H$ is diagonalizable with eigenvectors $H, X, Y$, so in fact any ideal of $\mathfrak{g}$ must be a direct sum of the corresponding eigenspaces; that is, it actually turns out to be true that any ideal must have as a basis some subset of $\{ H, X, Y \}$.
The physics CCR group is the Heisenberg group.
I'd follow WP, to fix concepts and notation, as you seem to confuse representations with the Lie algebras they represent.
The three-dimensional Lie algebra $\mathfrak h$ of the Heisenberg group H (over the real numbers) is known as the Heisenberg algebra. (Three generators, so three parameters, a,b,c.)
It may be represented using the space of 3×3 upper-triangular matrices of the form
$$\begin{pmatrix}
0 & a & c\\
0 & 0 & b\\
0 & 0 & 0\\
\end{pmatrix} , $$
with $a, b, c\in\mathbb R $; its exponential is the generic group element,
$$\begin{pmatrix}
1 & a & c+ab/2\\
0 & 1 & b\\
0 & 0 & 1\\
\end{pmatrix} , $$
Note that, in this (defining) representation, the algebra basis elements are not hermitean, and hence the group elements are not unitary. You'd be very wrong if you imagined the group is U(3).
The following three elements form a basis for $\mathfrak h$,
$$
X = \begin{pmatrix}
0 & 1 & 0\\
0 & 0 & 0\\
0 & 0 & 0\\
\end{pmatrix};\quad
Y = \begin{pmatrix}
0 & 0 & 0\\
0 & 0 & 1\\
0 & 0 & 0\\
\end{pmatrix};\quad
Z = \begin{pmatrix}
0 & 0 & 1\\
0 & 0 & 0\\
0 & 0 & 0\\
\end{pmatrix}. $$
The basis elements satisfy the commutation relations,
$$
[X, Y] = Z;\quad [X, Z] = 0;\quad [Y, Z] = 0. $$
In physics, it is also represented by infinite-dimensional hermitean matrices/operators,
$$\left[\hat x, \hat p\right] = i\hbar I;\quad \left[\hat x, \hbar I\right] = 0;\quad \left[\hat p, \hbar I\right] = 0. $$
Note the obligatory physics i upon transition to the hermitean basis! The group elements result from exponentiation with an i and a parameter, resulting into unitary infinite-dimensional matrices.
Further note in this rep that the trace of the central element Z, proportional to the infinite-dimensional identity here, is not zero, even though it amounts to a commutator. This is a standard famously frequent question on the PSE, with a subtle limit resolution.
The unitarity of the representations or not is not a feature of the Group, but, instead a feature of the representation.
Finally, the identity is never a feature of the Lie algebra, but of the universal enveloping algebra. In this representation, it coincides with the center Z.
Best Answer
Something which none of the other answers mentioned is that, assuming OP means $J_0=\sigma_1,J_1=\sigma_2,J_2=\sigma_3$, and $J_\pm=J_1\pm iJ_2$ or something like that, neither is actually a basis of $\mathfrak{su}(2)$.
My personal opinion is that none areas of mathematics are as butchered by the physicist's usual lack of precision as representation theory is. It is not always a problem, but it's nontheless good to get it right sometimes.
So first of all, the Lie algebra $\mathfrak{su}(2)$ consists of traceless, antihermitian matrices. The Pauli matrices are hermitian. But for example, let's define $$ T_i=-\frac{i}{2}\sigma_i, $$ then $$ [T_i,T_j]=-\frac{1}{4}[\sigma_i,\sigma_j]=-\frac{1}{4}2i\epsilon_{ijk}\sigma_k=-\frac{i}{2}\epsilon_{ijk}\sigma_k=\epsilon_{ijk}T_k. $$ Then the system $T_1,T_2,T_3$ does provide a basis for $\mathfrak{su}(2)$.
Secondly, despite the involvement of matrices with complex entries, $\mathfrak{su}(2)$ is a real Lie algebra, because the antihermiticity condition is not invariant under multiplication with $i$.
If we allow multiplication of elements with $i$, we get the set of all traceless matrices, which is $\mathfrak{sl}(2,\mathbb C)$, which I'll be considering as a complex Lie algebra (taken this way, $\mathfrak{sl}(2,\mathbb C)$ is the "complexification" of $\mathfrak{su}(2)$).
Thus, if complex linear combinations are allowed, then $(T_1,T_2,T_3)$, $(J_0,J_1,J_2)$, $(J_0,J_ +,J_-)$ etc. are all valid generators of $\mathfrak{sl}(2,\mathbb C)$.
I am noting here that $\mathfrak{sl}(2,\mathbb C)$ can also be "decomplexified" to obtain a real Lie algebra of dimension 6. For example if $T_1,T_2,T_3$ are the three antihermitian matrices I have written above, then $\mathfrak{sl}(2,\mathbb C)_\mathbb R$ is a real Lie algebra of dimension 6 whose generators can be taken to be say $T_1,T_2,T_3,iT_1,iT_2,iT_3$.
In the physics literature, for unitary Lie algebras, the generators are often taken to be hermitian rather than antihermitian because quantum mechanics prefer hermitian operators, and in many cases, complexifications and decomplexifications are left implicit and unmentioned and people will just happily multiply by $i$ without giving a second thought. But it should be noted that most matrices that are called the generators of $\mathfrak{su}(2)$ cannot actually be taken to be generators of $\mathfrak{su}(2)$, but of its complexification $\mathfrak{sl}(2,\mathbb C)=\mathfrak{su}(2)_\mathbb C$. This is especially true for the ladder opeators $J_\pm$, as those involve complex linear combinations.