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This is true for your "simple" case. In fact, assuming that the details of my proof below goes through correctly, the following is true.

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. I'll state them without full proofs for now, so at least there is a partial argument, and will edit the post to extend the proofs (maybe there are references where these are already proven, but I don't know of any).

First, I'll refer to the following sets of functions.

• U be thet set of functions f(x)∈C^∞(ℝ) which vanish on x≤0 and are positive on x>0.
• 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.

Sketch proof: Choose any smooth function r:ℝ⁺→ℝ⁺ with r(x)≥1 for small x, and r(x)=0 for x≥1. For example, we can use r(x) = exp(2-1/(1-x)) for x≤1. Then, the idea is to choose a sequence of positive reals α_k→0 and set

$$ g(x)=x^{\sum_kr_k(x/\alpha_k)}={\rm exp}\left({\rm log}(x)\sum_kr_k(x/\alpha_k)\right)$$ 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. The n^th derivative is bounded by K_nx^-ng(x)Σ_k1_{x>αk}α_k^-n, for positive constants K_n. As long as α_k go to zero fast enough, this will be of size O(x_n), so g∈C^∞(ℝ).

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 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 f_i,j,k(x)=max{|(d^i+j/dxⁱdy^j)g(x,y)|:|y|≤k}∈V. details to follow...