We can consider of a broad family of inellipses that includes Steiner's and Mandart's.
Given $\triangle ABC$ with $A=(0,0)$, $B=(c,0)$, $C=(b\cos A, b \sin A)$, and define point $P$ by
$$P = \frac{\frac1\alpha A + \frac1\beta B + \frac1\gamma C}{\frac1\alpha+\frac1\beta+\frac1\gamma} \tag{1}$$
(cumbersome reciprocals now make for cleaner expressions later). Let the cevians through $P$ meet the opposite sides in $A'$, $B'$, $C'$; that is, define
$$A' := \overleftrightarrow{AP}\cap\overleftrightarrow{BC} \qquad
B' := \overleftrightarrow{BP}\cap\overleftrightarrow{CA} \qquad
C' := \overleftrightarrow{CP}\cap\overleftrightarrow{AB}$$
One can show that there is a unique ellipse tangent to the side-lines of the triangle at $A'$, $B'$, $C'$. In the notation of my previous answer, it has equation
$$K_{20}\,x^2 + K_{11}\,x y + K_{02}\,y^2 + K_{10}\,x+K_{01}\,y+K_{00} = 0$$
with
$$\begin{align}
K_{20} &= \phantom{-}4\, (\alpha + \beta)^2 |\triangle ABC|^2 \\[4pt]
K_{11} &= \phantom{-}4 c \left((\alpha + \beta) (a \alpha \cos B - b \beta \cos A) + (\alpha - \beta) c \gamma \,\right) |\triangle ABC| \\[4pt]
K_{02} &= -4 (\alpha + \beta)^2 |\triangle ABC|^2 \\
&\phantom{=}\;+c^2 \left(
a^2 \alpha^2 + b^2 \beta^2 + c^2\gamma^2 + 2 bc\beta \gamma \cos A +
2 ca\gamma \alpha \cos B + 2 ab \alpha \beta \cos C
\right)\\[4pt]
K_{10} &= -8 c \alpha (\alpha + \beta) |\triangle ABC|^2 \\[4pt]
K_{01} &= -4 c^2 \alpha ( a \alpha \cos B - b \beta \cos A + c \gamma ) |\triangle ABC| \\[4pt]
K_{00} &= \phantom{-}4 c^2 \alpha^2 |\triangle ABC|^2
\end{align}$$
From these we get (after dividing-through by a common factor of $c^4$)
$$\begin{align}
p &= \left(
a^2 \alpha^2 + b^2 \beta^2 + c^2\gamma^2 + 2 bc\beta \gamma \cos A +
2 ca\gamma \alpha \cos B + 2 ab \alpha \beta \cos C
\right)^2 \\
q
&= -64 \alpha \beta \gamma (\alpha + \beta + \gamma )|\triangle ABC|^2
\end{align} \tag{2}$$
These, in conjunction with equations $(4.x')$ in my previous answer give conditions under which the inellipse is perfectly-packable.
We recover the Steiner inellipse case (again in my previous answer) by taking $\alpha=\beta=\gamma=1$ (and dividing-through by a common factor of $4$).
If we consider inellipses specifically of equilateral triangles, with $a=b=c=1$, then we have
$$\begin{align}
p &= \left(
\alpha^2 + \beta^2 + \gamma^2 + \beta \gamma +
\gamma \alpha + \alpha \beta
\right)^2 \\
q
&= -12 \alpha \beta \gamma (\alpha + \beta + \gamma )
\end{align} \tag{3}$$
For the particular case of ellipse centered on the triangle's line of symmetry (so that, say, $\alpha=\beta$) and packed with $3$ circles, equation $(4.3')$ of my previous answer reduces to
$$(3 \alpha^2 - 8 \alpha\gamma - 4 \gamma^2) (12 \alpha^2 - 2 \alpha\gamma - \gamma^2) = 0$$
which we can solve to get (ignoring negative values)
$$\gamma = \alpha \left(-1+\sqrt{13}\right) \qquad \gamma = \frac{\alpha}{2} \left(-2 +\sqrt{7}\right)$$
These correspond to the inellipses shown in @g.kov's answer.
Interestingly, all six equations $(4.x')$ factor when $\alpha=\beta$.
Ceva Ratios. If we define
$$\delta := \frac{|A'C|}{|BA'|} \qquad
\epsilon := \frac{|B'A|}{|CB'|} \qquad
\phi := \frac{|C'B|}{|AC'|}$$
Ceva's Theorem states that cevians $\overleftrightarrow{AA'}$, $\overleftrightarrow{BB'}$, $\overleftrightarrow{CC'}$ concur if and only if
$\delta\epsilon\phi = 1$. The point of concurrence can be written in the form $(1)$ with
$$\alpha:\beta:\gamma \;=\; 1 : \phi : \frac1\epsilon
\qquad \left( = \delta^0 : \phi^1 : \epsilon^{-1} \right) \tag{4}$$
(The relation $\delta\epsilon\phi=1$ means that $(4)$ can be written in a variety of ways, none of which is satisfyingly symmetric. However, I think of $(4)$ as being written "relative to" vertex $A$. The parameter $\alpha$ is naturally assigned its corresponding Ceva ratio, $\delta$, raised to the $0$-th power; but $\beta$ steals $\gamma$'s Ceva ratio, $\phi$, which is raised to the $1$-th power (because it's "looking forward"); likewise, $\gamma$ gets $\beta$'s Ceva ratio, $\epsilon$, which is raised to the $(-1)$-th power (because it's "looking backward").)
Some ellipses, such as Mandart's, may be easier to describe using Ceva ratios $\delta:\epsilon:\phi$ than reciprocal-barycentric coordinates $\alpha:\beta:\gamma$.
Actual Examples! Here's a family of the first six perfectly-packed ellipses in the equilateral triangle whose major axes align with a side of the triangle. Since the concurrence point $P$ lies on an axis of symmetry, so we have
$$\alpha:\beta:\gamma = 1 : \kappa: \kappa$$
for values of $\kappa$ we give below.
First, a family portrait ...
... and now the individual cases:
$$\begin{array}{ccc}
n = 1 & \qquad & n = 2 \\
\kappa = 1 & &
\kappa = \frac16\left(1 + \sqrt{7}\right)= 0.607\ldots
\end{array}$$
$$\begin{array}{ccc}
n = 3 & \qquad & n = 4 \\
\kappa = \frac1{12}\left(1 + \sqrt{13}\right) = 0.383\ldots & &
\kappa = 0.275\ldots \\
& & 1 + 4 \kappa - 20 \kappa^2 - 48 \kappa^3 + 72 \kappa^4 = 0
\end{array}$$
$$\begin{array}{ccc}
n = 5 & \qquad & n = 6 \\
\kappa = 0.213\ldots & &
\kappa = 0.173\ldots \\
1 + 4 \kappa - 32 \kappa^2 - 72 \kappa^3 + 144 \kappa^4 = 0 & &
1 + 4 \kappa - 44 \kappa^2 - 96 \kappa^3 + 144 \kappa^4 = 0
\end{array}$$
Best Answer
Consider an ellipse with semimajor axis $a$ and semiminor axis $b$ centered at $(0,0)$, where $a\geq b>0$. The eccentricity $e$ of this ellipse is given by $e=\sqrt{1-\frac{b^2}{a^2}}$. Without loss of generality, let $S=(+c,0)$ and $S'=(-c,0)$ be the foci of this ellipse, where $c:=ae$. Note that $PS+PS'=2a$ for every point $P$ on the ellipse. Thus, if $\big(h(\theta),k(\theta)\big)$ is the incenter of $P=\big(a\,\cos(\theta),b\,\sin(\theta)\big)$ for $\theta\in[0,2\pi)$, then, as you have found out, $$h(\theta)=\frac{2a^2e\,\cos(\theta)-ae\,PS+ae\,PS'}{2ae+2a}=\frac{e}{e+1}\Big(a\cos(\theta)-\frac{1}{2}PS+\frac{1}{2}PS'\Big)\,.$$ and $$k(\theta)=\frac{2abe\,\sin(\theta)}{2ae+2a}=\frac{e}{e+1}\big(b\,\sin(\theta)\big)\,.$$ Using $e=\sqrt{1-\frac{b^2}{a^2}}$ with $$PS=\sqrt{\big(a\,\cos(\theta)-ae)^2+b^2\sin^2(\theta)}\;\,\text{ and }PS'=\sqrt{\big(a\,\cos(\theta)+ae)^2+b^2\sin^2(\theta)}\;,$$ it can be easily seen that $PS=a\big(1-e\,\cos(\theta)\big)$ and $PS'=a\big(1+e\,\cos(\theta)\big)$. Therefore, $$h(\theta)=\frac{e}{e+1}a(1+e)\,\cos(\theta)=ae\,\cos(\theta)\,.$$ Let $A:=ae$ and $B:=\frac{e}{e+1}b$. Then, $$h(\theta)=A\,\cos(\theta)\text{ and }k(\theta)=B\,\sin(\theta)\,.$$ Therefore, the locus of the incenter of $PSS'$ is indeed an ellipse with the semimajor axis $A$ and the semiminor axis $B$. If $E$ is its eccentricity, then $$E=\sqrt{1-\frac{B^2}{A^2}}=\sqrt{1-\frac{b^2/a^2}{(e+1)^2}}=\sqrt{1-\frac{1-e^2}{(e+1)^2}}=\sqrt{1-\frac{1-e}{e+1}}=\sqrt{\frac{2e}{e+1}}\,.$$