$\newcommand\ZZ{\mathbb{Z}}\newcommand\RR{\mathbb{R}}$I'll explain how this computation should be done, and I'll show that there is at least one more closed path, with reflection sequence $(1,2,3,1,2,4)$.
I'll leave systematic search to those with better coding skills than I have.
Let $T$ be the tetrahedron, let $e_i$ be its vertices, let $F_i$ be the face of $T$ opposite $e_i$, and let $H_i$ be the affine $2$-plane containing $F_i$.
Given a sequence $(i_1, i_2, \ldots, i_m)$, we want to figure out whether there is a closed billiard path $\gamma$ which bounces off the walls in order $(F_{i_1}, F_{i_2}, \dots, F_{i_m})$.
If such a path $\gamma$ exists, I'll describe how to find it.
We extend $(i_1, i_2, \ldots, i_m)$ to a doubly infinite sequence, periodic modulo $m$.
Let $x_k$ be the point where $\gamma$ hits the face $F_{i_k}$, let $\gamma_k$ be the line segment from $x_k$ to $x_{k+1}$, let $L_k$ be the affine line spanned by $\gamma_k$ and let $v_k$ be a vector in the direction of $\gamma_k$. So $L_k = x_k + \RR v_k$ and $x_{k+1} = v_k + t_k x_k$ for some $t_k \in \RR$.
Let $s_i$ be the reflection over the wall $F_i$, so $s_i$ is a Euclidean, affine linear, transformation of $\RR^3$.
So we have $s_k x_k = x_k$ and $s_k v_{k-1} = v_k$. If we have candidate values of $x_0$ and $v_0$, then we can trace the rest of the path as follows: Given $(x_k, v_k)$, compute $t_k$ such that $x_{k} + t_{k} v_{k}$ lies in $H_{i_{k+1}}$, then $x_{k+1} = x_{k} + t_{k} v_{k}$. Meanwhile, $v_{k+1} = s_{i_{k+1}} v_k$.
The thing that can go wrong is that the point $x_{k+1}$ might lie in the $2$-plane $H_{i_{k+1}}$, but not in the triangle $F_{i_{k+1}}$ itself. So, as we go, we need to check that each $x_k$ that we compute has nonnegative coordinates.
We can use linear algebra to find out where $x_1$ and $v_0$ should be. Put $w = s_{i_m} \cdots s_{i_2} s_{i_1}$, so $w$ is a Euclidean transformation. If $m$ is even, then $w$ is probably a screw displacement; more on this later.
We must have $v_m = s_{i_m} s_{i_{m-1}} \cdots s_{i_1} v_0 = w v_0$ but, by periodicity, $v_0 = v_m$. Similarly, $L_0 = L_m = w L_0$. So, if $w$ is a screw displacement, then $v_0$ must point in the direction of the axis of $w$, and $L_0$ must be the axis of $w$. The point $x_1$ can be found as $L_0 \cap F_{i_1}$.
In order to find $v_0$ and $L_0$, it is convenient to embed the tetrahedron $T$ into $\RR^4$, putting the vertices $e_i$ at the standard basis vectors. So $T$ lies in the affine plane $\langle \delta,\ \rangle = 1$, where $\delta = e_1 + e_2 + e_3 + e_4$. Each affine linear transformation of $\langle \delta,\ \rangle = 1$ extends to a unique linear transformation of $\RR^4$.
Specifically, the reflection $s_4$ fixes $e_1$, $e_2$, $e_3$ and maps $e_4$ to $\tfrac{2}{3} e_1 + \tfrac{2}{3} e_2 + \tfrac{2}{3} e_3 - e_4 = \tfrac{2}{3} \delta - \tfrac{5}{3} e_4$. So $s_4$ is represented by the matrix $1 + (\tfrac{2}{3} \delta - \tfrac{8}{3} e_4) e_4^T$ and, similar, $s_i = 1 + (\tfrac{2}{3} \delta - \tfrac{8}{3} e_4) e_4^T$.
So $w = s_{i_m} s_{i_{m-1}} \cdots s_{i_1}$ is some $4 \times 4$ matrix. Then $v_0$ is an $1$-eigenvector of $w$ (which lies in $\delta^{\perp}$) and the $2$-plane spanned by $L_0$ is the generalized $1$-eigenspace of $w$. To get the point $x_0$, we intersect $L_0$ with the affine $2$ plane where $\langle \delta, \ \rangle = 1$ and where the $i_0$-th coordinate is $0$.
As an example, take the sequence $(1,2,3,1,2,4)$. I compute that
$$w = s_4 s_2 s_1 s_3 s_2 s_1 =
\frac{1}{729} \begin{bmatrix}
985 & 240 & 954 & 486 \\
880 & 825 & 90 & 486 \\
400 & 1104 & 1035 & 486 \\
-1536 & -1440 & -1350 & -729 \\
\end{bmatrix} . $$
I compute that the $1$-eigenspace of $w$ is spanned by $$v_0 := (-3/6, -1/6, -2/6, 6/6)^T.$$
Note that $\delta \cdot v_0 = 0$, as we'd expect.
I compute that the $1$-eigenspace of $w$ is spanned by $v_0$ and $(9/50, 9/50, 32/50, 0)^T$. I've normalized the second point to have $\langle \delta, \ \rangle = 1$ and to have $4$-th coordinate equal to $0$, so it lies in $H_{i_0} = H_4$. As it turns out, the other coordinates are nonnegative, so this point actually lies in the triangle $F_4$, and it is the point $x_0$.
We now compute $t_0$ such that $x_0 + t_0 v_0$ has $i_1$-th coordinate equal to $1$: We get $t_0 = 9/25$ and $x_1 = (0, 3/25, 13/25, 9/25)^T.$ Then $v_1 = s_1 v_0 = (3/6, -3/6, -4/6, 4/6)^T$. Continuing in this manner: $$(t_0, t_1, t_2, t_3, t_4, t_5) = (9/25,6/25,9/25,9/25,6/25,9/25)$$ and we have, $x_2=(3/25, 0, 9/25, 13/25)^T$, $x_3 = (9/50, 9/50, 0, 32/50)^T$, $x_4 = (0, 3/25, 9/25, 13/25)^T$, $x_5 = (3/25, 0, 13/25, 9/25)^T$ and $x_6 = x_0 = (9/50, 9/50, 32/50, 0)^T$. All the $x_i$ coordinates are nonnegative so we have found a closed path.
Some theoretical observations about the transformation $w$. Euclidean transformations of $\RR^3$ fall into eight classes: The orientation preserving ones (which occur for $m$ even) are the identity, rotation, translation and screw displacement. The orientation reversing ones (which occur for $m$ odd) are reflection, glide reflection, improper rotation and inversion in a point.
Assume that $i_j \neq i_{j+1}$ for all $j$ (including that $i_m \neq i_1$), as would be the case for any true reflection path, and assume that $m \geq 2$. Then I claim that $w$ cannot be the identity, a translation, a reflection, a glide reflection, or inversion in a point.
Consider the integer matrix $W:=3^m w = \prod_{j=m}^1 (3 + (2 \delta - 8 e_{i_j}) e_{i_j}^T)$.
So $W \equiv \prod_{j=m}^1 ((e_{i_j}-\delta) e_{i_j}^T) \bmod 3$.
Writing this out, it is $(e_{i_m} - \delta) e_{i_m}^T \cdots (e_{i_2} - \delta) e_{i_2}^T (e_{i_1} - \delta) e_{i_1}^T$. Now, since $i_j \neq i_{j+1}$, we have $e_{i_{j+1}}^T (e_{i_j} - \delta) = (-1)$,
$$W \equiv (-1)^{m-1} (e_{i_m} - \delta) e_{i_1}^T \bmod 3$$
and thus $\text{Tr}(W) \equiv (-1)^m \bmod 3$ (using that $i_1 \neq i_m$). So the trace of $W$ is an integer not divisible by $3$, and thus the trace of $w$ is not an integer. This rules out the identity, a translation, a reflection, a glide reflection, or inversion, whose traces are $4$, $4$, $2$, $2$, $-2$ respectively. (We also rule out rotations and glide rotations with angle $180^{\circ}$, whose trace is $0$.)
In particular, this means that there are no closed paths of odd length, since $w$ must be an improper rotation in that case, and those do not have a line which is mapped to itself by translation.
When $m$ is even, our options are a rotation or a screw displacement. We can distinguish them according to the Jordan forms of $w$ on the generalized $1$-eigenspace: A rotation corresponds to $\left[ \begin{smallmatrix} 1&0 \\ 0&1 \end{smallmatrix} \right]$ and a screw displacement corresponds to $\left[ \begin{smallmatrix} 1&1 \\ 0&1 \end{smallmatrix} \right]$. There doesn't seem to be any simple characterization of when we get which one, although screw displacements appear to be more common. The screw displacement has a line which is mapped to itself by translation; the rotation doesn't.
One final observation: The dihedral angles of the tetrahedron are
$\cos^{-1} (1/3) \approx 0.39 \pi > \pi/3$. So no closed path can contain a sequence of four bounces that only hit two walls. Combined with the fact that we should never use the same wall twice in a row, we should be able to cut down the search space to something tractable.
