Let's make the question a little more precise. We fix $n\in\mathbb{N}$ and $\varepsilon>0$, and define
$$\mathcal{A}_{n,\varepsilon}:=\bigcup_{k=1}^\infty\left\{\sum_{i=1}^ka_ix_i:x_i\in(0,\infty)^n\;\&\;a_i\geq -\varepsilon\;\forall i\right\}$$
and
$$\mathcal{B}_{n,\varepsilon}:=\bigcup_{k=1}^\infty\left\{\sum_{i=1}^ka_ix_i:x_i\in(0,\infty)^n\;\&\;a_i\geq -\varepsilon\;\forall i,\;\sum_{i=1}^ka_i=1\right\}.$$
Now we wish to characterize these sets in simpler terms.
If $\varepsilon<0$ then clearly
$$\mathcal{A}_{n,\varepsilon}=\mathcal{B}_{n,\varepsilon}=(0,\infty)^n$$
and if $\varepsilon=0$ then
$$\mathcal{A}_{n,0}=[0,\infty)^n$$
and
$$\mathcal{B}_{n,0}=(0,\infty)^n.$$
The case $\varepsilon>0$ is harder to see, and so for the sake of simplicity we work with $n=2$. Let $x=(x(1),x(2))\in\mathbb{R}^2$ and set $c=|x(1)|\vee|x(2)|$. Then
$$-\varepsilon\leq \frac{\varepsilon}{c}x(i)=:a_i.$$
We can choose $y_2(1)>0$ small enough so that
$$y_1(1):=\frac{c}{\varepsilon }-\frac{x(2)y_2(1)}{x(1)}>0$$
Then we can choose $y_1(2)>0$ small enough so that
$$y_2(2):=\frac{c}{\varepsilon}-\frac{x(1)y_1(2)}{x(2)}>0$$
This yields
$$\frac{\varepsilon}{c}[x(1)y_1(i)+x(2)y_2(i)]=x(i),\;\;\;i=1,2$$
and hence
$$a_1y_1+a_2y_2=x,\;\;\;y_i\in\mathbb{R}_+^2,\;\;\;a_i\geq -\varepsilon.$$
It follows that
$$\mathcal{A}_{2,\varepsilon}=\mathbb{R}^2.$$
The computations will be harder, but it should be relatively routine to show that $\mathcal{A}_{n,\varepsilon}=\mathbb{R}^n$ for all $n\in\mathbb{N}$.
Working with the convexity condition changes things. Consider when $n=1$. Then $\mathcal{B}_{1,\varepsilon}=\mathbb{R}$ when $\varepsilon>0$. Indeed, if $x>0$ then $1\cdot x=x$ does the trick. If $x\leq 0$ then we set $y=\frac{x-1-\varepsilon}{-\varepsilon}>0$ so that $-\varepsilon y+(1+\varepsilon)\cdot 1=x$.
I'll have to get back to you on $\mathcal{B}_{n,\varepsilon}$ when $\varepsilon>0$ for general $n\in\mathbb{N}$.
