How to calculate the concentration of all relevant species in a buffer of a given pH?
The value given for the pH of the solution seems to be incorrect*.

As the reaction is $\ce{HA &lt;=&gt; H+ + A-}$, your initial assumption that $[\ce{H+}]=[\ce{A-}]$ is correct.

From your chart, 

\begin{array}{l|ccc}
               &amp; \ce{HA} &amp;  \ce{H+}  &amp;  \ce{A-} \\\hline
\text{Initial} &amp; 0.25    &amp;  0        &amp;  0       \\
\text{Change}  &amp;  -x     &amp;  +x       &amp;  +x      \\
\text{End}     &amp; 0.25 -x &amp;  +x       &amp; +x       \\\hline
\end{array}

it follows that 
$$K_a=\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} = \frac{x^2}{0.25 -x} = 1.74\cdot10^{-5}.$$

We can solve this for $x$:
\begin{align}
x^2 &amp;= 1.74\cdot 10^{-5}(0.25-x)\\
x^2 + 1.74\cdot10^{-5} x - 4.35\cdot10^{-6} &amp;= 0\\
x &amp;= 0.00207698~\mathrm{M}\\
\end{align}

Thus, $[\ce{HA}] = 0.25~\mathrm{M} - 0.00207698~\mathrm{M} = 0.24792302~\mathrm{M}$ and $[\ce{H+}]=[\ce{A-}] = 0.00207698~\mathrm{M}$.

Also note that in your second attempt, the value $[\ce{A-}] = 4.35~\mathrm{M}$ makes no sense, as the amount of $\ce{A-}$ present should be in a 1:1 ratio to the amount of $\ce{HA}$ consumed in the reaction, and $[\ce{HA}] &lt; 4.35~\mathrm{M}$.

The given $\ce{pH}$ is likely incorrect because the concentration of $[\ce{H+}]$ ions for the given molarity and concentration does not match with the given $\ce{pH}$, and because the given $K_a$ is close to values of $K_a$ provided online.