Why no gravity lens around planets?

Contrary to the other answers, I will point out that gravitational lensing due to planets in the solar system is a significant and measurable effect. The measured positions of stars, as seen from a point in the solar system near the Earth are altered by the gravitational deflection due to the fields of the Sun and then, in order of decreasing effect, Earth Jupiter, Saturn, Venus, Uranus, Neptune.

The size of the effect depends on the angular separation of a star from the solar system object and can be anything from 0 to 70 microarcseconds.

This sounds very small, but is easily within the precision of positional measurements by the Gaia spacecraft. In fact, the effects of these deflections have to be taken into account in order to provide accurate positions as intermediate data for the calculation of the parallaxes and proper motions of stars.

A brief information sheet on the topic has been produced by ESA.

However, this lensing and these kinds of deflections are not strong enough to produce multiple images or Einstein rings. The angular size of a perfect Einstein ring is given by $$theta_E simeq left[ left(frac{4GM}{c^2}right) frac{D_{LS}}{D_L D_S}right]^{1/2} ,$$ where $M$ is the mass of the lens, $D_L$ is the distance from observer to lens, $D_S$ is the distance to the source and $D_{LS}$ is the distance between the lens and the source. Since in this case $D_{S} gg D_{L}$ and $D_{S} simeq D_{LS}$, then $$theta_E simeq left(frac{4GM}{c^2D_L}right)^{1/2} = 0.071 left(frac{M}{M_{rm Earth}}right)^{1/2} left( frac{D_L}{1{rm AU}}right)^{-1/2} {rm arcsec}$$

Taking Jupiter as an example, this yields $theta_E sim 0.6$ arcsec. But this is a lot smaller than the angular diameter of the planet ($sim 50$ arcsec). In other words, to see multiple images or rings you would need to be much further away from Jupiter than a few AU and a star that is directly behind Jupiter is completely hidden when viewed from the Earth.