Calculation of commutator of Lie algebra for affine linear maps

Question

This problem was answered before, but I'm stack with a technical point. Let $G$ be the Lie group of linear polynomials under composition (that is, affine transformations), $$\{x \mapsto ax+b, a\neq 0, a,b\in\mathbb{R}\}.$$ I'm going to use a dirty method, that is, without embedding the group to $GL(\mathbb{R},2)$. If we set $\gamma_1(t)=x+t$ and $\gamma_x(t)=e^tx$ Then, we can see $\{1,x\}$ is the basis of $g$, and $\gamma$'s are one-parameter subgroups. Given the formula $$\exp(x)\exp(y)\exp(-x)\exp(-y)=\exp([x,y]+\cdots)$$ I want to calculate $[x,1]$. So, by substituting $y=1$ and noting that $\exp(x)=\gamma_x(1)=ex$ and $\exp(1)=\gamma_1(1)=x+1$, we get $$\exp([x,1]+\cdots)=(ex)\circ(x+1)\circ(e^{-1}x)\circ(x-1)=x+e-1$$ However, the the constant term of the left side is 1. Why did this happen?