The fitted values are given by $\bar{y}_{x_i=t}$ for those observations with $x_i=t$, $t=0,1$. That is, the specific group means:
It is well-known (see e.g. here) that the general expression for the slope coefficient in a simple linear regression is
$$
\hat{\beta}_1 = \frac{\sum_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})}{\sum_{i=1}^n(x_i-\bar{x})^2}
$$
Here, the regressor $x_i$ is a dummy that takes the values $\{0,1\}$. Let $n_t$ denote the number of observations such that $x_i=t$, so that $n=n_0+n_1$. Then, $\bar{x}=n_1/n$. By this rule, write the numerator of the estimator as
$$
\sum_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})=\sum_{i=1}^nx_iy_i-n\bar{y}\frac{n_1}{n}=n_1(\bar{y}_{x_i=1}-\bar{y}),
$$
as multiplying by $x_i$ just "switches on" those $y_i$ for which $x_i=1$, and as the sum is the same as number of observations times average.
Similarly, we obtain for the numerator that
$$
\sum_{i=1}^n(x_i-\bar{x})^2=\sum_{i=1}^nx_i^2-n\frac{n_1^2}{n^2}=n_1-\frac{n_1^2}{n}=\frac{n_0n_1}{n},
$$
where it was used that squaring $x_i$ of course still gives either zeros or ones and that $n=n_0+n_1$ gives $nn_1-n_1^2=n_0n_1$. Putting things together gives
\begin{eqnarray*}
\hat{\beta}_1 &=& \frac{nn_1(\bar{y}_{x_i=1}-\bar{y})}{n_0n_1}\\
&=& \frac{n(\bar{y}_{x_i=1}-\bar{y})}{n_0}
\end{eqnarray*}
where $\bar{y}_{x_i=t}$ denotes the sample mean of the $y_i$ belonging to the observations for which $x_i=t$. Now, write
$$
\bar{y}=\frac{1}{n_0+n_1}(n_0\bar{y}_{x_i=0}+n_1\bar{y}_{x_i=1})
$$
Hence,
\begin{eqnarray*}
\hat{\beta}_1 &=& \frac{(n_0+n_1)(\bar{y}_{x_i=1}-\frac{1}{n_0+n_1}(n_0\bar{y}_{x_i=0}+n_1\bar{y}_{x_i=1}))}{n_0}\\
&=& \frac{(n_0+n_1)\bar{y}_{x_i=1}-n_0\bar{y}_{x_i=0}-n_1\bar{y}_{x_i=1}}{n_0}\\
&=& \frac{n_0\bar{y}_{x_i=1}-n_0\bar{y}_{x_i=0}}{n_0}\\
&=& \bar{y}_{x_i=1}-\bar{y}_{x_i=0}
\end{eqnarray*}
The intercept is, in general, given by
$$
\hat\beta_0=\bar{y}-\hat{\beta}_1\bar{x}
$$
Thus, writing $\sum_iy_i$ as the sum of group averages times their respective size,
\begin{eqnarray*}
\hat\beta_0&=&\frac{1}{n}(n_0\bar{y}_{x_i=0}+n_1\bar{y}_{x_i=1})-(\bar{y}_{x_i=1}-\bar{y}_{x_i=0})\frac{n_1}{n}\\
&=&\frac{n_0+n_1}{n}\bar{y}_{x_i=0}=\bar{y}_{x_i=0}
\end{eqnarray*}
Thus, the fitted values for $x_i=t$ are given by
$$\hat{y}=\hat\beta_0+\hat\beta_1\cdot t=\bar{y}_{x_i=0}+(\bar{y}_{x_i=1}-\bar{y}_{x_i=0})t,
$$
i.e.
$$
\hat{y}=\bar{y}_{x_i=0}
$$
when $x_i=0$ and
$$
\hat{y}=\bar{y}_{x_i=1}
$$
when $x_i=1$.
