Are there any other distributions whose shapes can be determined by just the mean and the standard deviation? For the normal distribution, With the mean, and the standard deviation, we know the shape of the specific normal distribution - that is, the center of it and how spread out it is. I wonder if there are other types of distributions whose shapes can be ascertained by just knowing the mean and the standard deviation. 
 A: There are many distributions defined by only two parameters. For example, the binomial distribution is defined by probability $p$ of success and number of independent trials $n$. Knowing the mean and standard deviation of a given binomial distribution, you can solve for $p$ and $n$ because you have two equations with two unknowns. Once you have solved for $n$ and $p$ then you have uniquely identified a binomial distribution.
Let $X\sim$binom($n,p$).
$E[X] = np$
$Var(X) = np(1-p)$
is a system of two equations with two unknowns $n$ and $p$ which can easily be solved for.
You can do this similar exercise for any distribution defined by only two parameters. 
A: The answer depends partly on what you mean by 'shape'. You could mean "the position of the distribution function at each $x$", or you could mean "the shape up to location and scale shifts". I'll take it to be the first (though in the case of the second, my answer under "2." relates pretty directly to it). I'll also assume you mean to refer only to the univariate case. 
1. Many distributions with two parameters can be determined by just their mean and standard deviation (almost any in common use). 
One could list many examples, but let's just consider two: the (two parameter) beta distribution and the lognormal. 
If you specify the population mean and the standard deviation, you can write the two parameters of either distribution in terms of them. 
a) Beta: $f(x;\alpha,\beta) = \frac{x^{\alpha-1}(1-x)^{\beta-1}} {\text{B}(\alpha,\beta)}\,, 0<x<1; \alpha,\beta>0$
Mean = $\frac{\alpha}{\alpha+\beta}$ $\qquad$ ($=\mu$, say)
Variance = $\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}$ $\qquad$ ($=\sigma^2$, say)
and of course the standard deviation, $\sigma$, is just the square root of that variance.
Then:
${\alpha} = \mu \left(\frac{\mu (1 - \mu)}{\sigma^2} - 1 \right),$ and
${\beta} = (1-\mu) \left(\frac{\mu (1 - \mu)}{\sigma^2} - 1 \right).$
In some cases you may not be able to write a closed form algebraic expression for the parameters in terms of $\mu$ and $\sigma$ 
Indeed, that this can generally be done (combined with the law of large numbers) is why method of moments is as broadly useful as it is. In the two-parameter case, one can replace the mean and variance by their sample estimates (being a simple reparameterization of $(E(X), E(X^2))$) and solve equations relating the parameters to the mean and variance like those above (either algebraically or numerically) to obtain parameter estimates.
b) Lognormal:
Here the density is usually written in terms of parameters $\mu$ and $\sigma$ which don't represent mean and variance of the lognormal, so here I'll call the mean and variance of the lognormal $m$ and $v$ (breaking the usual convention of using Greek letters).
Mean: $e^{\mu+\sigma^2/2} = m$
Variance: $(e^{\mu+\sigma^2/2})^2(e^{\sigma^2}-1) = v$
Hence:
$\sigma=\sqrt{\log(1+\frac{v}{m^2})}$
$\mu = \log(m)-\frac{_1}{^2}\log(1+\frac{v}{m^2})$

2. You can also generate families for which this can be done quite easily - take any fixed distribution (if it has parameters you can simply fix them at whichever specific values you like), and then embed it in a location-scale family via a linear transformation involving two parameters which are linearly related to the mean and standard deviation.
For example, take $X$ distributed as a standard uniform. It has mean $\frac{1}{2}$ and standard deviation $\frac{1}{\sqrt{12}}$. 
If $Y=\theta+\phi X$, then the mean and standard deviation of $Y$ will just be $\theta+\phi/2$ and $\phi\cdot \frac{1}{\sqrt{12}}$. 
Hence, for $Y$ we can solve for $\theta$ and $\phi$ immediately.
(Edit: this is the same as the situation whuber is discussing in his comment)
