Suppose we construct the torus, $T$, as a CW-complex in the following way: Given a wedge sum of two circles that is generated by the letters $a$ and $b$, we attach a $2$-cell, $e_1^2$, on the wedge sum by the word $aba^{-1}b^{-1}$. We would like to prove that every point on the torus constructed in this manner has a neighbourhood that is homeomorphic to an open set in $\mathbb{R}^2$.
Start Of Proof:
Let $p \in T$. We have two cases:
$p \notin \partial e_1^2$
- Then we can find a neighbourhood $U$ small enough so that $U \subset \text{Int}(e_1^2)$. Then $U$ is trivially homeomorphic to an open set of $\mathbb{R}^2$ since $e_1^2$ has the subspace topology of $\mathbb{R}^2$.
$p \in \partial e_1^2$
- $\ldots$
I want to express the idea that if $p \in \partial e_1^2$, then an open set around $p$ (that is not all of $e_1^2$) will consist of open arcs along $\partial e_1^2$ and open sets in $\text{Int}(e_1^2)$ "spilling out" of $\partial e_1^2$, which altogether is still an open set homeomorphic to $\mathbb{R}^2$. I don't know how to make this argument rigorous, however.
Unfortunately being a manifold is very special property of a topological space, and certainly not even every CW complex satisfies the criteria to be one. You will not get what you are looking for using only formal properties of CW structures - and certainly what you describe is a generic set of CW attaching data. Clearly we need to be able to identify some extra structure available in the torus, which we can exploit to show that it is locally Euclidean. This will most likely arrive in the form of an explicit description of the characteristic map, constructed only using the data of the attaching map that you describe. I've tried to this below, and shall leave you to judge whether it is useful to you or not.
Consider the product $S^1\times S^1$. This is obviously the torus we are interested in, but for our purposes let us momentarily relive ourselves of that knowledge. We'll call this space $S^1\times S^1$, and we'll call the more abstract complex constructued with your attaching data $X$.
Let $I=[0,1]$ be the unit interval and define
$$\Phi:I^2=I\times I\rightarrow S^1\times S^1,\qquad \Phi(s,t)=(e^{2\pi is},e^{2\pi it}).$$
Note that $\partial I^2\cong I\times\{0,1\}\cup\{0,1\}\times I$ and
$$\Phi(\partial I^2)=S^1\vee S^1\subset S^1\times S^1$$
where we view the wedge $S^1\vee S^1$ as the subset of the product consisting of those points with at least one coordinate equal to the basepoint $\ast=(1,0)$. Thus there is a commutative diagram $\require{AMScd}$ \begin{CD} \partial I^2@>>> \ I^2\\ @V\Phi| V V @VV \Phi V\qquad(\ast)\\ S^1\vee S^1 @>>> S^1\times S^1. \end{CD} where the horizontal maps are the obvious inclusions. Now it is not difficult to see that $\Phi$ maps $I^2\setminus\partial I^2$ injectively onto $S^1\times S^1\setminus S^1\vee S^1$, so what we have constructued above is a relative homeomorphism $\Phi:(I^2,\partial I^2)\xrightarrow{\cong}(S^1\times S^1,S^1\vee S^1)$. On the other hand there is a relative homeomorphism $\alpha:(D^2,S^1)\xrightarrow{\cong} (I^2,\partial I^2)$ where
$$\alpha|_{S^1}(e^{i2\pi t})=\begin{cases}(4t,0)&0\leq t\leq \frac{1}{4}\\(1,4t-1)&\frac{1}{4}\leq t\leq\frac{1}{2}\\(3-4t,1)&\frac{1}{2}\leq t\leq \frac{3}{4}\\(0,4-4t)&\frac{3}{4}\leq t\leq 1\end{cases}$$
and it is clear that the composition $\Phi|_{\partial I^2}\circ\alpha|_{S^1}:S^1\rightarrow S^1\vee S^1$ exactly describes the attaching map you have written down above.
Thus the relative homeomorphism given by the diagram labelled $(\ast)$ is a cell structure for the torus $T^2$ that you have described and we conclude that there is a homeomorphism
$$X\cong S^1\times S^1.$$
Thus the torus $X$ has the structure of a smooth manifold and in particular is locally Euclidean.
Now since this last step is a bit of a shortcut, let us forget that $S^1\times S^1$ is smooth. Let us explicitly describe an open neighbourhood of a point $(x,y)\in S^1\vee S^1\subseteq S^1\times S^1$ in the boundary of top cell. It's easiest to consider two cases, so lets start with the case that $(s,t)$ is not the basepoint, and so is of the form $(x,(1,0))$ or $((1,0),y)$. Both subcases work out to be more or less the same so let's be concrete and assume this point is $(e^{2\pi it},(1,0))$ for some $t\in(0,1)$. Then if $0<\epsilon<t$, consider the open ball $B_\epsilon(t,0)\subseteq\mathbb{R}^2$. Extend the domain of $\Phi$ to all of $\mathbb{R}^2$ in the obvious way and consider the set $U=\Phi^{-1}(\Phi(B_\epsilon(t,0))\subseteq \mathbb{R}^2$. It is not difficult to see that $U$ is a union of disjoint open balls that each map homeomorphically onto an open neighbourhood of $(e^{2\pi it},(1,0))\in S^1\times S^1$. This is the neighbourhood you are looking for. The cases for other possible points in $S^1\vee S^1$ follos similarly, and I'll leave these up to you.