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I can show that this is true for your "simple" case.

If g(x,y) ∈ C^∞(ℝ²) vanishes on x ≤ 0 then it decomposes as g(x,y) = a(x)G(x,y) where a(x) ∈ C^∞(ℝ) vanishes on x ≤ 0 and G(x,y) ∈ C^∞(ℝ²).

This can be shown by proving the statements below. They could possibly be standard results, but I've never seen them before. First, I'll refer to the following sets of functions.

• Let U be thet set of functions f(x)∈C^∞(ℝ) which vanish on x≤0 and are positive on x>0.
• Let V be the set of functions f:ℝ⁺→ℝ such that x^-nf(x)→0 as x→0, for each positive integer n.

The statements I need to show the main result are as follows.

Lemma 1: For any f∈V, there is a g∈U such that f(x)/g(x)→0 as x→0.

Proof: Choose any smooth function r:ℝ⁺→ℝ⁺ with 1≥r≥0, r(x)>0 for small x, and r(x)=0 for x≥1. For example, we can use r(x) = exp(-1/(1-x)) for x≤1. Then, the idea is to choose a sequence of positive reals α_k→0 satisfying ∑_kα_k < ∞, and set

$$g(x) = x^{\theta(x)},\ \ \ \theta(x)=\sum_{k=1}^\infty r(x/\alpha_k)$$

for x>0 and g(x)=0 for x≤0. Only finitely many terms in the summation will be nonzero outside any neighborhood of 0, so it is a well defined expression, and smooth on x>0. Clearly, θ(x)→∞ and, therefore, x^-ng(x)→0 as x→0. It needs to be shown that all the derivatives of g vanish at 0 so that g∈U. As r and all its derivatives are bounded with compact support, r⁽ⁿ⁾(x) ≤K_nx^-n-1 for some constants K_n. The n^th derivative of θ is

$$\theta^{(n)}(x)=\sum_k\alpha_k^{-n}r^{(n)}(x/\alpha_k)\le K_nx^{-n-1}\sum_k\alpha_k$$

which has polynomially bounded growth in 1/x. The derivatives of log(g) satisfy

$$\frac{d^n}{dx^n}\log(g(x))=\frac{d^n}{dx^n}\left(\log(x)\theta(x)\right)$$

which also has polynomially bounded growth in 1/x. However, the derivative on the left hand side is g⁽ⁿ⁾(x)/g(x) plus a polynomial in g⁽ⁱ⁾(x)/g(x) for i<n. So, induction gives that g⁽ⁿ⁾(x)/g(x) has polynomially bounded growth in 1/x and, multiplying by g(x), g⁽ⁿ⁾(x)→0 as x→0.

By definition of f∈V, there is a decreasing sequence of positive reals ε_k such that f(x)≤xⁿ for x≤ε_n. We just need to make sure that α_k≤ε_n+1 for k≥n to ensure that g(x)≥x^n-1 for ε_n+1≤x≤min(ε_n,1). Then f(x)/g(x) goes to zero at rate x as x→0.

Lemma 2: For any sequence f₁,f₂,...∈V there is a g∈U such that f_k(x)/g(x)→0 as x→0 for all k.

Sketch Proof: The idea is to apply Lemma 1 to f(x)=Σ_kλ_k|f_k(x)| for positive reals λ_k. This works as long as f∈V, which is the case if Σ_kλ_kmin(x,1)^-ksup_x≤k|f_k(x)| is finite, and this condition is easy to ensure.

Lemma 3: For any sequence f₁,f₂,...∈V there is a g∈U such that f_k(x)/g(x)ⁿ→0 as x→0 for all positive integers k,n.

Proof: Apply Lemma 2 to the doubly indexed sequence f_k,n=|f_k|^1/n.

The result follows from applying lemma 3 to the triply indexed sequence f_i,j,k(x)=max{|(d^i+j/dxⁱdy^j)g(x,y)|:|y|≤k}∈V. Then, there is an a∈U such that f_ijk(x)/a(x)ⁿ→0 as x→0. Set G(x,y) = f(x,y)/a(x) for x>0 and G(x,y) = 0 for x ≤ 0. On any bounded region for x>0, the derivatives of G(x,y) to any order is a sum of terms, each of which is a product of f_ijk(x,y)/a(x)ⁿ with derivatives of a(x), so this vanishes as x→0. Therefore, G∈C^∞(ℝ²).