As for the Fibonacci sequence, let us find a basis.

 1. If $a_{0,j} = 0$, then $a_{ij} = \binom{i+j-1}{j}$.
 2. If $a_{0,j} = 1$, then $a_{ij} = \binom{i+j}{j}$.
 3. If $a_{0,j} = j+1$, then $a_{ij} = \binom{i+j+1}{j}$.
 4. If $a_{0,j} = \frac{(j+1)(j+2)}2$, then $a_{ij} = \binom{i+j+2}{j}$.

Any quadratic polynomial $a+bj+cj^2$ is an affine combination of $0$, $1$, $j+1$, $\frac{(j+1)(j+2)}2$. The answer will be the combination of $\binom{i+j-1}{j}$, $\binom{i+j}{j}$, $\binom{i+j+1}{j}$, $\binom{i+j+2}{j}$ with the same coefficients.

If I have solved the linear system correctly, one gets
$$a_{ij} = (1-a)\binom{i+j-1}{j} + (a-b+c) \binom{i+j}{j} + (b-3c) \binom{i+j+1}{j} + 2c \binom{i+j+2}{j}$$

Note that for $i=j=0$ this formula yields $a_{00} = a$, which was not well-defined in the question (but does not participate in the recurrence and is therefore irrelevant).