Interesting problem. One should distinguish between two distinct cases, according as whether $p_1=p_2$ or not.
The case $p_1=p_2$ (let's call this common prime $p$) is easy. You can just subtract the two equations and get:
$$(a_1-a_2)x + (b_1-b_2) \equiv 0 \pmod p $$
which gives either a unique choice for $x$ (if $a_1\neq a_2$) or none at all (if $a_1=a_2$ but the $b_i$'s are distinct). In the former case, you can easily check whether that coordinate $x$ corresponds to an actual point on the elliptic curves; for randomly chosen curves, the chance of this happening is about 1/2. Note however that standard curve parameters often mandate that $a=-3$, in which case you are always in the latter case, and intersections simply don't occur for curves over the same base field.
The case $p_1\neq p_2$ is more subtle, because it isn't obvious how to interpret the question anymore. Mathematically speaking, the problem doesn't really make sense, because the fields $\mathbb{F}_{p_1}$ and $\mathbb{F}_{p_2}$ are distinct, unrelated objects.
One can look, however, for points $(x,y)$ with integer coefficients (or even, in some sense, rational coefficients), with the property that when reduced modulo $p_1$ we get a point on the first curve, and when reduced modulo $p_2$ we get a point on the second curve. Constructing such points is straightforward: take any point $P_1$ on the first curve, any point $P_2$ on the second curve, and apply the Chinese remainder theorem to their coefficients to get what you want. But this may not be the problem you are actually interested in, because if you do that, the bit size of the coefficients of the resulting point $(x,y)$ will be as large as the sum of the bit sizes of $p_1$ and $p_2$ in general.
One could also ask for such integer solutions $(x,y)$ with the additional requirement that, for example, $x$ and $y$ should be at most $\min(p_1,p_2)$ in absolute value (or perhaps $\max(p_1,p_2)$ depending on the application). This amounts to looking for points with small coefficients on the lifted curve
$$y^2 \equiv x^3 + \hat{a}x + \hat{b} \pmod {p_1p_2}$$
where $\hat{a},\hat{b}$ are obtained by Chinese remaindering. Heuristically, if $p_1$ and $p_2$ are roughly of the same size, say $m$ bits, such a point should exist with good probability (there are about $2^m$ possible $x$ coordinates of the right size, and at least one of the corresponding $y$ coordinates fall in the right interval with probability around $2^{-m}$; one can make this rough estimate more precise if necessary). It is not completely obvious how to find such a point, however. The usual technique to find small solutions of algebraic equations is to use some variant of Coppersmith's theorem, but depending on the precise version of the problem you are interested in, the bounds offered by Coppersmith are likely to be insufficient.
If you can say more about your precise application, one could have a closer look. It might even be worth writing a paper about it.
