Classical test theory: Covariance of two variables
I do not myself use classical test theory, but from a statistical point of view the answer to this should be straightforward. The answer is that you are right, and that the first solution is wrong. 

An early indication that the first formula is incorrect is that a :
$$
Cov(\tau_{x},\tau_{y}) = 0
$$
because $\tau_{x}$ and $\tau_{y}$ are constants (I assume $\tau_{x}$ and $\tau_{y}$ are constant in your question because your equations state that they are expected values, which are always constant). The formula above holds because constants can't covary with one another. 

To show you are the one who is correct, we need to dig a little bit deeper. One basic property of covariance is, for $X$, $Y$ random variables and $a$, $b$ constants:
$$
Cov(X+a, Y+b) = Cov(X,Y)
$$
Intuitively, adding a constant to a random variable doesn't affect its linear relationship with another random variable. 

In your particular example, if I understand correctly, $\epsilon_{x}$, $\epsilon_{y}$ are random variables, and $\tau_{x}$, $\tau_{y}$ are fixed constants. Plugging these into the above formula, 
$$
Cov(\epsilon_{x} + \tau_{x}, \epsilon_{y} + \tau_{y}) = Cov(\epsilon_{x}, \epsilon_{y})
$$
And of course, using the notation in your question, $X$ = $\tau_{x}$ + $\epsilon_{x}$= $\epsilon_{x}$ + $\tau_{x}$ and $Y$ = $\tau_{y}$ + $\epsilon_{y}$= $\epsilon_{y}$ + $\tau_{y}$. Therefore, using the variables from your question,
$$
Cov(X,Y) = Cov(\epsilon_{x}, \epsilon_{y})
$$