Solution:
The only such pair is $(0,0)$, which clearly works. To prove this is the only one, let us prove an auxiliary result first.

Lemma If $\alpha, \beta, \gamma$ are reals so that $\alpha+\beta+\gamma=0$ and $|\alpha|,|\beta|,|\gamma| \geq 2$, then
$$|\alpha \beta+\beta \gamma+\gamma \alpha|<|\alpha \beta \gamma|$$
Proof. Some two of these reals have the same sign; WLOG, suppose $\alpha \beta>0$. Then $\gamma=-(\alpha+\beta)$, so by substituting this,
$$|\alpha \beta+\beta \gamma+\gamma \alpha|=\left|\alpha^{2}+\beta^{2}+\alpha \beta\right|,\ |\alpha \beta \gamma|=|\alpha \beta(\alpha+\beta)|$$
So we simply need to show $|\alpha \beta(\alpha+\beta)|>\left|\alpha^{2}+\beta^{2}+\alpha \beta\right|$. Since $|\alpha| \geq 2$ and $|\beta| \geq 2$, we have
$$\begin{aligned} & |\alpha \beta(\alpha+\beta)|=|\alpha||\beta(\alpha+\beta)| \geq 2|\beta(\alpha+\beta)| \\ & |\alpha \beta(\alpha+\beta)|=|\beta||\alpha(\alpha+\beta)| \geq 2|\alpha(\alpha+\beta)| . \end{aligned}$$
Adding these and using triangle inequality,
$$\begin{aligned} 2|\alpha \beta(\alpha+\beta)| & \geq 2|\beta(\alpha+\beta)|+2|\alpha(\alpha+\beta)| \geq 2|\beta(\alpha+\beta)+\alpha(\alpha+\beta)| \\ & \geq 2\left(\alpha^{2}+\beta^{2}+2 \alpha \beta\right)>2\left(\alpha^{2}+\beta^{2}+\alpha \beta\right) \\ & =2\left|\alpha^{2}+\beta^{2}+\alpha \beta\right| \end{aligned}$$
Here we have used the fact that $\alpha^{2}+\beta^{2}+2 \alpha \beta=(\alpha+\beta)^{2}$ and $\alpha^{2}+\beta^{2}+\alpha \beta=\left(\alpha+\frac{\beta}{2}\right)^{2}+\frac{3 \beta^{2}}{4}$ are both nonnegative. This proves our claim.

For our main problem, suppose the roots of $x^{3}+a x+b$ are the integers $r_{1}, r_{2}, r_{3}$ and the roots of $x^{3}+b x+a$ are the integers $s_{1}, s_{2}, s_{3}$. By Vieta's relations, we have
$$\begin{gathered} r_{1}+r_{2}+r_{3}=0=s_{1}+s_{2}+s_{3} \\ r_{1} r_{2}+r_{2} r_{3}+r_{3} r_{1}=a=-s_{1} s_{2} s_{3} \\ s_{1} s_{2}+s_{2} s_{3}+s_{3} s_{1}=b=-r_{1} r_{2} r_{3} \end{gathered}$$
If all six of these roots had an absolute value of at least 2, by our lemma, we would have
$$|b|=\left|s_{1} s_{2}+s_{2} s_{3}+s_{3} s_{1}\right|<\left|s_{1} s_{2} s_{3}\right|=\left|r_{1} r_{2}+r_{2} r_{3}+r_{3} r_{1}\right|<\left|r_{1} r_{2} r_{3}\right|=|b|$$
which is absurd. Thus at least one of them is in the set $\{0,1,-1\}$; WLOG, suppose it's $r_{1}$.

1. If $r_{1}=0$, then $r_{2}=-r_{3}$, so $b=0$. Then the roots of $x^{3}+b x+a=x^{3}+a$ are precisely the cube roots of $-a$, and these are all real only for $a=0$. Thus $(a, b)=(0,0)$, which is a solution.

2. If $r_{1}= \pm 1$, then $\pm 1 \pm a+b=0$, so $a$ and $b$ can't both be even. If $a=-s_{1} s_{2} s_{3}$ is odd, then $s_{1}, s_{2}, s_{3}$ are all odd, so $s_{1}+s_{2}+s_{3}$ cannot be zero. Similarly, if $b$ is odd, we get a contradiction.

The proof is now complete.

Solution:
The only such pair is $(0,0)$, which clearly works. Let us prove this is the only one. In what follows, we use $\nu_{2}(n)$ to denote the largest integer $k$ so that $2^{k} \mid n$ for any non-zero $n \in \mathbb{Z}$.

If one of the cubics has 0 as a root, say the first one, then $0^{3}+0 \cdot a+b=0$, so $b=0$. Then the roots of $x^{3}+b x+a=x^{3}+a$ are precisely the cube roots of $-a$, and these are all real only for $a=0$. Thus $(a, b)=(0,0)$.

So suppose none of the roots are zero. Take the cubic $x^{3}+a x+b$, and suppose its roots are $x, y, z$. We cannot have $\nu_{2}(x)=\nu_{2}(y)=\nu_{2}(z)$; indeed, if we had $x=2^{k} x_{1}, y=2^{k} y_{1}, z=2^{k} z_{1}$ for odd $x_{1}, y_{1}, z_{1}$, then
$$0=x+y+z=2^{k}\left(x_{1}+y_{1}+z_{1}\right)$$
But $x_{1}+y_{1}+z_{1}$ is odd, and hence non-zero, so this cannot happen.

Thus we can assume WLOG that $\nu_{2}(x)>\nu_{2}(y)$. Then the third root is $-(x+y)$. Similarly, the three roots of $x^{3}+b x+a$ can be written as $p, q,-(p+q)$ where $\nu_{2}(p)>\nu_{2}(q)$. By Vieta's relations,
$$\begin{gathered} x y-x(x+y)-y(x+y)=-\left(x^{2}+x y+y^{2}\right)=a=p q(p+q) \\ p q-p(p+q)-q(p+q)=-\left(p^{2}+p q+q^{2}\right)=b=x y(x+y) \end{gathered}$$
Suppose $x=2^{k} x_{1}$ and $y=2^{\ell} y_{1}$ for odd $x_{1}, y_{1}$ and $k>\ell$; in particular $k>0$. Then
$$x y(x+y)=2^{k} x_{1} \cdot 2^{\ell} y_{1} \cdot\left(2^{k} x_{1}+2^{\ell} y_{1}\right)=2^{k+2 \ell} x_{1} y_{1}\left(2^{k-\ell} x_{1}+y_{1}\right)$$
Here $x_{1} y_{1}\left(2^{k-\ell} x_{1}+y_{1}\right)$ is clearly odd, so $\nu_{2}(x y(x+y))=k+2 \ell$.

Also,
$$x^{2}+x y+y^{2}=2^{2 k} x_{1}^{2}+2^{k} x_{1} \cdot 2^{\ell} y_{1}+2^{2 \ell} y_{1}^{2}=2^{2 \ell}\left(2^{2 k-2 \ell} x_{1}^{2}+2^{k-\ell} x_{1} y_{1}+y_{1}^{2}\right)$$
Again, all the terms in the second factor are even except $y_{1}^{2}$, so the entire factor is odd. This means $\nu_{2}\left(x^{2}+x y+y^{2}\right)=2 \ell$.

Therefore
$$\nu_{2}(x y(x+y))>\nu_{2}\left(x^{2}+x y+y^{2}\right)$$
Similarly, one may show
$$\nu_{2}(p q(p+q))>\nu_{2}\left(p^{2}+p q+q^{2}\right)$$
But then
$$\nu_{2}(b)=\nu_{2}(x y(x+y))>\nu_{2}\left(x^{2}+x y+y^{2}\right)=\nu_{2}(p q(p+q))>\nu_{2}\left(p^{2}+p q+q^{2}\right)=\nu_{2}(b)$$
Here we have used the fact that $\nu_{2}(n)=\nu_{2}(-n)$ for any integer $n$. But this is a contradiction, proving our claim.