Is the posterior always a compromise between the prior and the data? Suppose that we are interested in learning the proportion of the population theta with a particular property (for instance, the fraction of the population who are male).

Step 1
Nice question! Sadly it is not true. I'll talk in terms of using coin flips to determine the bias $\mathrm{\theta <!--\; \theta \; -->}$ of a weighted coin, where $\mathrm{\theta <!--\; \theta \; -->}$ is the probability that the coin flips heads. Here's the idea: consider a prior in which we assign very high probability to values of $\mathrm{\theta <!--\; \theta \; -->}$ very close to either 0 or 1 and very low probability otherwise, such that the prior expectation is $\frac{1}{2}$. Now suppose we see, say, $\frac{2}{3}$ heads in the sample. This is very unlikely if $\mathrm{\theta <!--\; \theta \; -->}$ is close to 0 and much more likely if it's close to 1, so the posterior distribution is concentrated on values of $\mathrm{\theta <!--\; \theta \; -->}$ close to 1 and in particular the posterior mean is close to 1, which makes it potentially larger than either the prior expectation $\frac{1}{2}$ or the sample fraction $\frac{2}{3}$.
You can see hints of this already in your beta distribution calculation: the inequalities you want assume that $\mathrm{\alpha <!--\; \alpha \; -->}$ and $\mathrm{\beta <!--\; \beta \; -->}$ are positive, and are false for, say, $\mathrm{\alpha <!--\; \alpha \; -->}=0,\mathrm{\beta <!--\; \beta \; -->}=-<!--\; -\; -->\frac{1}{2}$. Of course the beta integral does not converge in this case but this gives us an idea of what to look for.
Formally, take H (for "height") to be a large positive constant and w (for "width") and $\mathrm{ϵ<!--\; ϵ\; -->}$ to be small positive constants, and consider the "triangular" prior with probability density function
$p(\mathrm{\theta <!--\; \theta \; -->})=\{\begin{array}{ll}H(1-<!--\; -\; -->\frac{\mathrm{\theta <!--\; \theta \; -->}}{w})+\mathrm{ϵ<!--\; ϵ\; -->}& \text{ if }0\le <!--\; \le \; -->\mathrm{\theta <!--\; \theta \; -->}\le <!--\; \le \; -->w\\ \mathrm{ϵ<!--\; ϵ\; -->}& \text{ if }w\le <!--\; \le \; -->\mathrm{\theta <!--\; \theta \; -->}\le <!--\; \le \; -->1-<!--\; -\; -->w\\ H(1-<!--\; -\; -->\frac{1-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}}{w})+\mathrm{ϵ<!--\; ϵ\; -->}& \text{ if }1-<!--\; -\; -->w\le <!--\; \le \; -->\mathrm{\theta <!--\; \theta \; -->}\le <!--\; \le \; -->1.\end{array}.$
We have ${\int <!--\; \int \; -->}_0^{1}p(\mathrm{\theta <!--\; \theta \; -->})d\mathrm{\theta <!--\; \theta \; -->}=Hw+\mathrm{ϵ<!--\; ϵ\; -->}$ so this is a pdf as long as $Hw+\mathrm{ϵ<!--\; ϵ\; -->}=1$. It's symmetric about $\frac{1}{2}$, so the prior expectation $\mathbb{E}(\mathrm{\theta <!--\; \theta \; -->})$ is $\frac{1}{2}$.
Step 2
Now suppose we flip 3 coins and 2 of them are heads. Then the posterior density is the normalization of ${\mathrm{\theta <!--\; \theta \; -->}}^2(1-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->})p(\mathrm{\theta <!--\; \theta \; -->})$, and so the posterior expectation is $\frac{\mathbb{E}({\mathrm{\theta <!--\; \theta \; -->}}^3(1-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}))}{\mathbb{E}({\mathrm{\theta <!--\; \theta \; -->}}^2(1-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}))}$. This is a slightly tedious but doable calculation which I will punt to WolframAlpha; we get
$\mathbb{E}({\mathrm{\theta <!--\; \theta \; -->}}^2(1-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}))=(-<!--\; -\; -->\frac{w^3}{12}+\frac{w^2}{6})H+\frac{\mathrm{ϵ<!--\; ϵ\; -->}}{12}$
$\mathbb{E}({\mathrm{\theta <!--\; \theta \; -->}}^3(1-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}))=(-<!--\; -\; -->\frac{w^5}{15}+\frac{w^4}{5}-<!--\; -\; -->\frac{w^3}{4}+\frac{w^2}{6})H+\frac{\mathrm{ϵ<!--\; ϵ\; -->}}{20}$
so the posterior expectation is their quotient. This is a bit annoying to write out in full so let's just talk about how it behaves asymptotically. If w is small then the polynomials in w above are dominated by their terms with smallest exponent, namely $\frac{w^2}{6}$, which in both cases comes from the portion of the integral corresponding to $\mathrm{\theta <!--\; \theta \; -->}\approx <!--\; \approx \; -->1$ and hence ${\mathrm{\theta <!--\; \theta \; -->}}^n(1-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->})\approx <!--\; \approx \; -->1-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}$; importantly this portion of the integral approximately does not depend on the exponent n of $\mathrm{\theta <!--\; \theta \; -->}$. We have $H=\frac{1-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->}}{w}$ which gives, for both w and $\mathrm{ϵ<!--\; ϵ\; -->}$ small,
$\mathbb{E}({\mathrm{\theta <!--\; \theta \; -->}}^2(1-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}))=\frac{w}{6}+\frac{\mathrm{ϵ<!--\; ϵ\; -->}}{12}+O(w^2+\mathrm{ϵ<!--\; ϵ\; -->}w)$
$\mathbb{E}({\mathrm{\theta <!--\; \theta \; -->}}^3(1-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}))=\frac{w}{6}+\frac{\mathrm{ϵ<!--\; ϵ\; -->}}{20}+O(w^2+\mathrm{ϵ<!--\; ϵ\; -->}w)$
so we see that by taking w to be small but $\mathrm{ϵ<!--\; ϵ\; -->}$ to be much smaller we can arrange for the posterior expectation to be arbitrarily close to 1, and in particular not in the interval $[\frac{1}{2},\frac{2}{3}]$, as expected. To be concrete we can take, say, $w=0.01,\mathrm{ϵ<!--\; ϵ\; -->}=0.0001$.
If you're looking for a conceptual upshot, one conceptual upshot here is that when the prior is very "lopsided" like this, the prior mean is not a good summary of it, and when the prior is concentrated on two very different hypotheses, apparently small amounts of evidence can tilt the balance between them dramatically.