The real Matrix of change of basis. Not really. Only in $\mathbb{R}^n$. Suppose we have an $\mathbb{R}$-vector space $E$, $\text{dim}(E)=n$, and two bases $\alpha:=\{v_i\}$ and $\beta:=\{w_i\}$ of it. 
We can consider the maps to $\mathbb{R}^n$ given by the coordinates $v\mapsto [v]_\alpha$ and $v\mapsto[v]_\beta$. Let $I:E\rightarrow E$ be the identity function and let $[I]_{\alpha}^{\beta}$ be the matrix of $I$ in the bases $\alpha$ and $\beta$. We have then, from the construction of matrix of a linear transformation, that $$[v]_{\beta}=[I]_{\alpha}^{\beta}\cdot[v]_{\alpha}.$$
This matrix $[I]_{\alpha}^{\beta}$ is often called the matrix of change of coordinates (from $\alpha$ to $\beta$), and sometimes the matrix of change of basis (from $\alpha$ to $\beta$). But let us call it here only the matrix of change of coordinates (from $\alpha$ to $\beta$) since that is what is doing.
Consider now $T:E\rightarrow E$ the linear transformation defined by $T(v_i):=w_i$, for all $i$. This linear transformation is sending the vectors of the basis $\alpha$ to the vectors of the basis $\beta$. Let us call it the transformation of change of basis (because that is what is doing). 
Assume from this point on that $E=\mathbb{R}^n$. Then we have the standard basis $e:=\{e_i\}$, which has the property that coordinates in that basis look like the same vectors, i.e. $[v]_e=v$, for every $v\in E=\mathbb{R}^n$. Therefore the matrix $[T]_{e}^{e}$ of $T$ in the basis $e$ satisfies 
$$w_i=[w_i]_e=[T(v_i)]_e=[T]_{e}^{e}\cdot[v_i]_e=[T]_{e}^{e}v_i$$
Therefore, this matrix $[T]_{e}^{e}$ is sending the vectors of the basis $\alpha$ to the vectors of the basis $\beta$. We could call it matrix of change of basis since that is what is doing.

Questions:
  
  
*
  
*What is the relationship between $[T]_{e}^{e}$ and $[I]_{\alpha}^{\beta}$? 
  
*What is $[T]_{e}^{e}$ usually called?
  

Example:
Assume that $E=\mathbb{R}^2$, $\alpha=\{v_1=\begin{bmatrix}1\\1\end{bmatrix}, v_2=\begin{bmatrix}2\\1\end{bmatrix}\}$, and $\beta=\{w_1=\begin{bmatrix}2\\2\end{bmatrix}, w_2=\begin{bmatrix}3\\2\end{bmatrix}\}$.
Then we have 

$$\begin{align}v_1&=\phantom{-}\frac{1}{2}w_1+0w_2\\v_2&=-\frac{1}{2}w_1+1w_2\end{align}$$

Therefore 

$$[I]_{\alpha}^{\beta}=\begin{bmatrix}\frac{1}{2}&-\frac{1}{2}\\0&\phantom{-}1\end{bmatrix}$$

To compute $[T]_{e}^{e}$ we can evaluate $T(e_1)$ and $T(e_2)$, expand them in the basis $e$ and put the coefficients as columns in a matrix. Since $$\begin{align}e_1&=-1v_1+1v_2\\e_2&=2v_1-v_2\end{align}$$
we get that $$\begin{align}T(e_1)&=T(-v_1+v_2)&=-w_1+w_2\\T(e_2)&=T(2v_1-v_2)&=2w_1-w_2\end{align}$$
This tells us  that 

$$[T]_{e}^{e}=\begin{bmatrix}1&1\\0&2\end{bmatrix}.$$

Notice that when we compute the inverse of $[I]_{\alpha}^{\beta}$ in this case, we don't get $[T]_{e}^{e}$ as this answer seems to be implying. We get 
$$([I]_{\alpha}^{\beta})^{-1}=\begin{bmatrix}2&1\\0&1\end{bmatrix}$$
 A: (This is a much-revised answer). 
Doing my best to follow your notation, I'm going to say that
$$
[s]_B
$$
is the vector of coefficients $(c_1, \ldots, c_n)^t$ with the property that $c_1 b_1 + \ldots + c_n b_n = s$, where $B$ is the basis whose vectors are $b_1, b_2, \ldots$. 
I'm going to write $e_i$ for the $i$th standard basis vector in $R^n$ -- the one with all entries 0 except the $i$th, which is 1. This unfortunately means that 
$$
[e_i]_E = e_i
$$
where $E = e_1, e_2, \ldots$ is the standard basis. 
Let $V$ be a matrix such that 
$$
V[e_i]_E = [v_i]_E
$$
i.e., the $i$th column of $V$ contains the standard coordinates of $v_i$. In your example, 
$$
V = \begin{bmatrix} 1 & 2 \\ 1 & 1\end{bmatrix}.
$$
Define $W$ analogously. 
Then we have
\begin{align}
[e_i]_E &= V^{-1} [v_i]_E \text{ (1) }\\
[e_i]_E &= W^{-1} [w_i]_E \text{ (2) }\\
V[e_i]_E &=  [v_i]_E \text{ (3) }\\
W[e_i]_E &=  [w_i]_E \text{ (4) }\\
\end{align}
Your definition of $T_e^e$, is, in this notation, that it's the matrix such that
\begin{align}
 [w_i]_E &= T_e^e [v_i]_E \\
 W[e_i]_E &= T_e^e [v_i]_E \text{, by eq 4}\\
 W[e_i]_E &= T_e^e V[e_1]_E \text{, by eq 3}
\end{align}
Since this holds for each $i$, we can laminate together all the $[e_i]_E$ into the identity matrix to get
\begin{align}
 W I  &= T_e^e VI \text{, so that}\\
 W V^{-1} &= T_e&e.
\end{align}
$$\newcommand{\Iab}{I_\alpha^\beta}$$
Now let's look at $\Iab$.
First, look at $v = v_i$, the $i$th entry of the first basis. In the first basis, its coordinates are just $(1, 0, 0, \ldots)^t$, i.e., 
$$
[v_i]_\alpha = [e_i]_E
$$
That's a statement about equality of coordinate representations. 
Second, according to equation (1) above, we have
\begin{align}
[e_i]_E &= V^{-1} [v_i]_E 
\end{align}
Combining, we get
\begin{align}
[v_i]_\alpha &= V^{-1} [v_i]_E 
\end{align}
We can apply this to a linear combination of the $v_i$ to get 
\begin{align}
[v]_\alpha &= V^{-1} [v]_E \text{ (5) }
\end{align}
and correspondingly, 
\begin{align}
[v]_\beta &= W^{-1} [v]_E \text{ (6) }
\end{align}
Now it all falls into place. with substitution, we get 
\begin{align}
[v]_\beta &= \Iab [v]_\alpha \text{, from the definition of $\Iab$}\\
W^{-1} [v]_E &= \Iab [v]_\alpha \text{, applying Eq. (6)}\\
W^{-1} [v]_E &= \Iab V^{-1} [v]_E \text{, applying Eq. (5)}\\
W^{-1}  &= \Iab V^{-1}  \text{, because prev. eqn is true for all $v$}\\
W^{-1}V  &= \Iab 
\end{align}
Here's a matlab script verifying that these computations lead to the result you got when you worked it out by hand:
v1 = [1;1]
v2 = [2;1]
w1 = [2;2]
w2 = [3;2]
V = [v1, v2]
W = [w1, w2]
Tee = W * V^(-1)
Iab = (W^(-1)) * V

Summary: $\Iab = W^{-1} V$; $T_e^e = W V^{-1}$. 
And just to make the comment-stream below continue to make sense, I'll suggest that putting the $\alpha$ up and the $\beta$ down leads to a nice situation in which "up" and "down" indices appear to cancel. 
For your second question ("what is $T_e^e$ usually called?") I'd say "The matrix of the transformation taking the $v$s to the $w$s," at least in the case where the $v$s and $w$s are in $R^n$. That's the term I've heard used most often, anyhow. It may be that a poll of mathematicians would reveal something different. 
One last thought: when one of the two bases actually is the standard basis, then either $W$ or $V$ is the identity, and the two matrices $\Iab$ and $T_e^e$ end up being inverses. 
