We give two proofs, the first by manipulation of the integral, and a second much shorter one that uses probabilistic language.
Proof 1: Symmetry about $a$ means that $f(a+z)=f(a-z)$ for all $z$.
Suppose that the expectation $E(X)$ exists. Then
$$E(X)=\int_{-\infty}^\infty xf(x)\,dx.$$
Rewrite as
$$\int_{-\infty}^\infty \left(a +(x-a)\right)f(x)\,dx,$$
and then as
$$\int_{-\infty}^\infty \left(a +(x-a)\right)f(x)\,dx,$$
which is equal to
$$\int_{-\infty}^\infty af(x)\,dx+\int_{-\infty}^\infty (x-a)f(x)\,dx.\tag{1}$$
The first integral is $a$. We need to show the second integral is $0$. This is essentially obvious by symmetry: $(x-a)f(x)$ is an odd function. But we do the details.
In the second integral of (1), make the change of variable $y=x-a$. Then the integral becomes
$$\int_{-\infty}^\infty yf(a+y)\,dy.$$
Break up at $y=0$. We get
$$\int_{-\infty}^0 yf(a+y)\,dy +\int_0^\infty yf(a+y)\,dy.\tag{2}$$
For the first integral, make the change of variable $z=-y$. We get
$$\int_{z=\infty}^0 (-1)(-z)f(a-z)\,dz.$$
Using $f(a-z)=f(a+z)$, and some minor fooling with minus signs, we end up with
$$-\int_{z=0}^\infty zf(a+z)\,dz.$$
This cancels the second integral of (2).
Proof 2: A more probabilistic (and therefore better) way of doing the problem is to first let $X=a+Y$. Assume that $E(X)$ exists (it need not). Then $E(X)=a+E(Y)$. We need to show that $E(Y)=0$.
The density function $g(y)$ of $Y$ is symmetric about $y=0$. It follows that
that the random variable $Y$ has the same distribution as the random variable $-Y$. Thus $E(Y)=E(-Y)=-E(Y)$ and therefore each is $0$.
