Let's express a point \$\vec p\$ on your 3D ellipse as a parametric equation, using the vectors \$ \vec c\$ for the center, \$ \vec M \$ for the semi-major axis and \$ \vec m \$ for the semi-minor axis, and a parameter \$ 0 \le t \le 2 \pi\$:
$$\vec p(t) = \cos(t) \vec M + \sin(t) \vec m + \vec c$$
From this we can derive three points on your 2D ellipse, at the tip of each of the 3D ellipse's major axes, and one "diagonal" point between them:
$$ (M_x + c_x, M_y + c_y)\\
(m_x + c_x, m_y + c_y)\\
(d_x, d_y) = (\frac {M_x + m_x} {\sqrt{2}} + c_x, \frac {M_y + m_y} {\sqrt{2}} + c_y)$$
These will be three solutions to the general equation of an ellipse centered at a point \$\vec c\$:
$$ A (x - c_x)^2 + B (x - c_x) (y - c_y) + C (y - c_y)^2 = 1$$
We can clear out all instances of \$\vec c\$ to shift our problem to the origin for now (and add it back later). We can then express this equation and our three known solutions as a system of linear equations:
$$\begin{bmatrix}
M_x^2 & M_x M_y & M_y^2\\
m_x^2 & m_x m_y & m_y^2\\
d_x^2 & d_x d_y & d_y^2\\
\end{bmatrix}
\begin{bmatrix}
A\\B\\C
\end{bmatrix}
=
\begin{bmatrix}
1\\1\\1
\end{bmatrix}
$$
You can solve this system for the unknowns \$A\$, \$B\$, \$C\$ (eg. using Cramer's Rule or other solving techniques of your liking) to get the equation of the 2D ellipse.
Taking the formulae from Wikipedia, that leaves us with...
$$\begin{align}
a,b&=\frac {-{\sqrt {2{\Big (}(B^{2}-4AC){\Big )}\left((A+C)\pm {\sqrt {(A-C)^{2}+B^{2}}}\right)}}}{B^{2}-4AC}\\ \\
\theta &= \begin{cases}\arctan \left({\frac {1}{B}}\left(C-A-{\sqrt {(A-C)^{2}+B^{2}}}\right)\right)&{\text{for }}B\neq 0\\0&{\text{for }}B=0,\ A<C\\90^{\circ }&{\text{for }}B=0,\ A>C\end{cases}
\end{align}$$
...for the length of the semi-major axis \$a\$, semi-minor axis \$b\$, and the angle \$\theta\$ from the positive x axis to the major axis.