Are these proofs logically equivalent? Here are two proofs, firstly:
x = 0.999...
10x = 9.999...
    = 9 + 0.999...
    = 9 + x
9x = 9
x = 1

And secondly:
x = 1 - 1 + 1 - 1 + 1 - 1 ...
  = 1 - (1 - 1 + 1 - 1 + 1 ...
  = 1 - x
2x = 1
x = 1/2

The fourth line in the first and the third line in the second use the same trick. Does the legitimacy of the first proof guarantee that of the second?

No.  The first series is absolutely convergent, so its terms can be rearranged without changing the sum, while the second series is divergent.  
The number $0.\overline{9}$ can be expressed in the form 
\begin{align*}
0.\overline{9} & = \sum_{k = 1}^{\infty} \frac{9}{10^k}\\
               & = \frac{9}{10} \sum_{k = 0}^{\infty} \frac{1}{10^k}
\end{align*}
which is a convergent geometric series, so 
\begin{align*}
0.\overline{9} & = \frac{9}{10} \cdot \frac{1}{1 - \frac{1}{10}}\\
               & = \frac{9}{10} \cdot \frac{1}{\dfrac{9}{10}}\\
               & = \frac{9}{10} \cdot \frac{10}{9}\\
               & = 1
\end{align*}
In fact, the series 
$$0.\overline{9} = \sum_{k = 1}^{\infty} \frac{9}{10^k}$$
is absolutely convergent since the series 
$$\sum_{k = 1}^{\infty} \bigg| \frac{9}{10^K} \bigg| = \sum_{k = 1}^{\infty} \frac{9}{10^k}$$
converges.  Any rearrangement of an absolutely convergent series has the same sum.
The series 
$$1 - 1 + 1 - 1 + 1 - 1 + \cdots = \sum_{k = 0}^{\infty} (-1)^k$$
diverges since the sequence of partial sums $\{1, 0, 1, 0, 1, 0, \ldots\}$
alternates between $1$ and $0$.  Since the series never converges to a limit, it is meaningless to say that 
$$x = 1 - 1 + 1 - 1 + 1 - 1 + \cdots$$
A series must be absolutely convergent to guarantee that rearranging the terms of the series does not change the sum.
