In the scenario you describe least squares regression will allow you to tell a very straightforward story:
First of all, imagine that you have no dichotomous independent variable. So:
(1) $y_{i} = \beta_{0} + \beta_{1}x_{1i} + \varepsilon_{i}$
Your regression describes the relationship between your dependent variable $y$ and your continuous independent variable $x_{1}$ as a straight line, with intercept $\beta_{0}$ and slope $\beta_{1}$. Cool? Cool.
Now add both the dichotomous independent variable $x_{2}$ and the interaction between $x_{1}$ and $x_{2}$ to the model:
(2) $y_{i} = \beta_{0} + \beta_{1}x_{1i} + \beta_{2}x_{2i} + \beta_{3}x_{1i}x_{2i} + \varepsilon_{i}$
So now what is your model telling you? Well, (assuming $x_{2}$ is coded 0/1) when $x_{2} = 0$, then the model reduces to equation (1) because $\beta_{2} \times 0 = 0$ and $\beta_{3} \times x_{1} \times 0 = 0$. So that is easy-peasy puddin' pie.
What about when $x_{2} =1$? Well now the $y$-intercept is $\beta_{0} + \beta_{2}$ (Right? Because $\beta_{2} \times 1 = \beta_{2}$).
And the slope of the line relating $y$ to $x_{1}$ is now $\beta_{1} + \beta_{3}$ (Right? Because $\beta_{1}\times x_{1} + \beta_{3} \times x_{1} \times 1 = \beta_{1}\times x_{1} + \beta_{3} \times x_{1} = (\beta_{1} + \beta_{3})\times x_{1}$).
So when $x_{2}=1$ you simply have a second regression line relating $y$ to $x_{1}$, with a different intercept (if $\beta_{2} \ne 0$) and a different slope (if $\beta_{3} \ne 0$ which will be true if you tested a significant interaction term in, say, ANOVA).
How do you communicate this? A single graph with two regression lines overlaying your data (possibly with different colored/shaped/sized markers when $x_{2}=1$), and a label indicating which line corresponds to $x_{2}=0$ and $x_{2}=1$. Also providing your audience with the values of the $\beta$s and their standard errors and/or confidence intervals is good (like, in a table of multiple regression results).
Cool? Cool.
Finally, while all the above tells you about trend relationships between $y$ and $x_{1}$ given $x_{2}$, least squares regression also tells you about strength of association. If you had a single independent variable, you'd probably want to use something like $R^{2}$ to describe this strength of association, but when you add variables $R^{2}$ doesn't quite mean what it did before. So you might use generalized $R^{2}$, or Pseudo-$R^{2}$ or some such.
Best Answer
Not in a meaningful way, unless you have additional information to add to your model.
If all you know is the values of these categorical predictors, then any data belonging to the same category (or combination of categories) will obviously be predicted by your regression model to have the exact same value. To change that, you'd need to have some way of additionally distinguishing points within the same category. Barring that, I don't see how you could say something like "actually this point should be a bit higher than the others" (other than choosing randomly).
Smoothing already implies some additional dimension of interest in the data. That is, there has to be something to "smooth over". E.g. if you had time series data you might have reason to smooth your predictions over time, taking a weighted average of data from adjoining time points. (Although I'd say it generally makes more sense to add this dimension to your model explicitly at the fitting stage, rather than smooth predictions from a model without it.)
Update: Also, it's not necessarily a problem if your predictions are discrete. As Glen_b touched upon as well, a linear regression assumes the data follow a continuous (Normal) distribution around some mean (that is a function of your regressors), but that mean can be discrete, and in fact have any arbitrary distribution you fancy. So if you think your categorical regressors are a good (enough) model for your data, there's no reason to be concerned.