What is the expression for intensity of Laser beam focussed at a point at an angle theta?
Methodology
We can apply simple ray optics to this problem, and it gets easy!
What you're describing is an Orthographic Projection (OP).
So, what you'll do is

realize circles subject to OP become ellipses
take the center of the circles of equal intensity (in your case, $x^2+y^2=\text{const.}$), and project that. It'll be the center of all the ellipses of equal intensity.
figure out what the major and minor axes of the ellipses are going to be
simply write down the formula for points on ellipses with given axes.

Orthogonal Projection
For the math behind OP itself, I'd like to refer you to the wikipedia link given above. It's really just a single matrix multiplication, if your original (circular) image plane and the projected plane both contain the origin $(x,y,z)=(0,0,0)$; it becomes especially easy if the intersection of these two planes is e.g. the $y$-coordinate axis, which we can assume here; you can calculate the normal vector of the slanted plane simply by multiplying the cosine (or sine, depending on your construction) of the incident angle to unit vector on the $z$ axis, then.
Major axis of the ellipses of equal intensity
For the question of what the major axis is going to be: it's the projection of the diameter of the original circle that's parallel to the plane intersection (source). Luckily for you, you've defined that intersection to be a diameter of the circle! Since all points on the intersection of original and image plane are invariant to the OP, you have to do nothing to find the major axis of any circle of given intensity: it's simply the distance on the $y$-axis of the original!
Minor axis
The minor axis is simply perpendicular. You'll find that the projection is also especially easy, and is simply the &quot;complementary&quot; trigonometric function to what you've used to find the normal of the projection plane.
Overall formula
Now that we know how to convert perpendicular incidence to slanted incidence circles, we just have to figure out how to do the inverse. Luckily, since the projection is such a well-formed matrix, that's just taking the inverse– and that's trivial, given the structure of the matrix!