Yes, it's perfectly possible.
Start off with the obvious Ansatz,
$$
\psi(x)
=
\begin{cases}
A e^{\kappa x} & x<-a \\
B e^{\kappa x} + Ce^{-\kappa x} & -a<x<a \\
D e^{-\kappa x} & a<x
\end{cases}
$$
and then set up the boundary conditions,
\begin{align}
Ae^{-\kappa a} & = Be^{-\kappa a}+Ce^{\kappa a} \\
Be^{\kappa a}+Ce^{-\kappa a} & = De^{-\kappa a}
\end{align}
from the continuity of $\psi$ and
\begin{align}
B\kappa e^{-\kappa a}-C\kappa e^{\kappa a} - A\kappa e^{-\kappa a}
& = -\alpha Ae^{-\kappa a}-\alpha Be^{-\kappa a}-\alpha Ce^{\kappa a}
\\
-D\kappa e^{-\kappa a} - B \kappa e^{\kappa a} + C \kappa e^{-\kappa a}
& =
-D \alpha e^{-\kappa a} - B\alpha e^{\kappa a} - C\alpha e^{-\kappa a}
\end{align}
from $\psi'(\pm a^+)-\psi'(\pm a^-)=-2\alpha \psi(\pm a)= -\alpha[\psi(\pm a^+)+\psi(\pm a^-)]$ (assuming $\hbar=m=1$), using the latter form to keep things symmetric.
Now, putting everything into a single system, you get
\begin{align}
& Ae^{-\kappa a} &- Be^{-\kappa a}&-Ce^{\kappa a}& & = 0 \\
& & Be^{\kappa a}&+Ce^{-\kappa a} &- De^{-\kappa a}& = 0 \\
& A(\alpha-\kappa) e^{-\kappa a}&+ B(\alpha+\kappa)e^{-\kappa a}&+C(\alpha-\kappa)e^{\kappa a} & & = 0 \\
& & B(\alpha-\kappa) e^{\kappa a} &+ C(\alpha+\kappa) e^{-\kappa a} &+ D (\alpha-\kappa) e^{-\kappa a} & = 0
\end{align}
which in matrix form reads
$$
\begin{pmatrix}
e^{-\kappa a} &- e^{-\kappa a}&-e^{\kappa a}& 0 \\
0 & e^{\kappa a}&e^{-\kappa a} &- e^{-\kappa a}\\
(\alpha-\kappa) e^{-\kappa a}& (\alpha+\kappa)e^{-\kappa a}&(\alpha-\kappa)e^{\kappa a} & 0 \\
0 & (\alpha-\kappa) e^{\kappa a} &(\alpha+\kappa) e^{-\kappa a} & (\alpha-\kappa) e^{-\kappa a}
\end{pmatrix}
\begin{pmatrix}
A \\ B \\ C \\ D
\end{pmatrix}
=0.
$$
This only has trivial zero solutions if the system is nonsingular, so you need
$$
\det\mathopen{}
\begin{pmatrix}
e^{-\kappa a} &- e^{-\kappa a}&-e^{\kappa a}& 0 \\
0 & e^{\kappa a}&e^{-\kappa a} &- e^{\kappa a}\\
(\alpha-\kappa) e^{-\kappa a}& (\alpha+\kappa)e^{-\kappa a}&(\alpha-\kappa)e^{\kappa a} & 0 \\
0 & (\alpha-\kappa) e^{\kappa a} &(\alpha+\kappa) e^{-\kappa a} & (\alpha-\kappa) e^{-\kappa a}
\end{pmatrix}
\mathclose{}
=0,
\tag 1
$$
and this is the core of your quantization condition. From there, it's up to you to massage this expression down to whatever your specific goal is.
That said, it's definitely worth connecting this brute-force result with what you would have gotten if you had banked on a definite even or odd symmetry from the outset. To do that, note that for parity $\pm 1$, you can obtain your vector of coefficients from a smaller one by means of a rectangular matrix,
$$
\begin{pmatrix}
A \\ B \\ C \\ D
\end{pmatrix}
=
\begin{pmatrix}
A \\ B \\ \pm B \\ D
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & \pm 1 & 0 \\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
A \\ B \\ D
\end{pmatrix}
,
\tag 2
$$
so you can just dot the matrix in $(1)$ with the rectangular matrix in $(2)$ (and just take the first three rows to trim to a 3-by-3 square) before taking the determinant.
Alternatively, if you want to go for broke, you can impose full even/odd symmetry into your vector of coefficients by imposing a symmetric mixing between $A$ and $D$, which in practice means replacing the rectangular matrix in $(2)$ with the full mixing matrix
$$
\begin{pmatrix}
1 & 0 & 0 & 1 \\
0 & 1 & 1 & 0 \\
0 & - 1 & 1 & 0\\
-1 & 0 & 0 & 1
\end{pmatrix}
.
$$
Once you've done that, again, the quantization condition is to take the determinant to zero,
$$
\det\mathopen{}\left(
\begin{pmatrix}
e^{-\kappa a} &- e^{-\kappa a}&-e^{\kappa a}& 0 \\
0 & e^{\kappa a}&e^{-\kappa a} &- e^{\kappa a}\\
(\alpha-\kappa) e^{-\kappa a}& (\alpha+\kappa)e^{-\kappa a}&(\alpha-\kappa)e^{\kappa a} & 0 \\
0 & (\alpha-\kappa) e^{\kappa a} &(\alpha+\kappa) e^{-\kappa a} & (\alpha-\kappa) e^{-\kappa a}
\end{pmatrix}
\begin{pmatrix}
1 & 0 & 0 & 1 \\
0 & 1 & 1 & 0 \\
0 & - 1 & 1 & 0\\
-1 & 0 & 0 & 1
\end{pmatrix}
\right)\mathclose{}
=0,
$$
and then there is a series of elementary row operations that can transform the matrix (without changing the determinant) into a block matrix, i.e. two blocks of two-by-two, with zeros on the off-block-diagonal. The full determinant factorizes into the determinants of each block, and those are the separate quantization conditions for the even and odd solutions.
(No promises regarding the detailed correctness of the formulas in this answer $-$ it's entirely possible that one or more algebraic errors snuck in. You need to replicate this work in full in any case.)
