If this is assumed to be associative then
$$
ij = jiij = 1
$$
and so
$$
-j = iij = i.
$$
But
$$
1 = -jj = ij = -j
$$
but also
$$
1 = -ii = ji = -i.
$$
Thus $1 = -1$ and your system is inconsistent.
Because the algebra is not associative it can't be the trivial algebra.
Note that the algebra is spanned by $1,i,j$.
Suppose now that $j = \alpha i$ for some scalar $\alpha$. Then
$$
-i = (\alpha i)i = -\alpha^2
$$
and $i = \alpha^2$, making the algebra trivial. This is impossible, so $i$ and $j$ are linearly independent.
Suppose now that $1 = \alpha i + \beta j$ for some scalars $\alpha,\beta$. Then
$$
1 = (\alpha i + \beta j)^2 = -\alpha^2 - \beta^2 - \alpha\beta(i + j).
$$
This implies that one of $\alpha, \beta$ must be nonzero. Thus
$$
1 = \frac{-\alpha\beta}{1+\alpha^2+\beta^2}(i+j).
$$
Because $i,j$ are linearly independent, this implies
$$
\alpha = \beta = \frac{-\alpha^2}{1+2\alpha^2} \implies 2\alpha^3 + \alpha^2 + \alpha = 0
$$
and because $\alpha$ cannot be zero we get
$$
2\alpha^2 + \alpha + 1 = 0.
$$
The discriminant of this quadratic equation is
$$
1 - 4(2)(1) < 0
$$
so there are no real solutions, which contradicts the assumption that $1$ is a linear combination of $i,j$.
Thus $1,i,j$ is a basis for the algebra, and we can define multiplication by
$$
(\alpha + \beta i + \gamma j)(\alpha' + \beta' i + \gamma' j) = \alpha\alpha' - \beta\beta' - \gamma\gamma' + (\alpha\beta' + \beta\alpha' - \gamma\beta')i + (\alpha\gamma' + \gamma\alpha' - \beta\gamma')j.$$
This multiplication satisfies the requisite identities; this proves that the algebra is consistent and unique.
