The definition of Determinant in the spirit of algebra and geometry

That seems quite opaque: It's a way of computing a quantity rather than telling what exactly it is or even motivating it. It also leaves completely open the question of why such a function exists and is well-defined. The properties you give are sufficient if you're trying to put a matrix in upper-triangular form, but what about other computations? It also gives no justification for one of the most important properties of the determinant, that $\det(ab) = \det a \det b$.

I think the best way to define the determinant is to introduce the wedge product $\Lambda^* V$ of a finite-dimensional space $V$. Given that, any map $f:V \to V$ induces a map $\bar{f}:\Lambda^n V \to \Lambda^n V$, where $n = \dim V$. But $\Lambda^n V$ is a $1$-dimensional space, so $\bar{f}$ is just multiplication by a scalar (independent of a choice of basis); that scalar is by definition exactly $\det f$. Then, for example, we get the condition that $\det f\not = 0$ iff $f$ is an isomorphism for free: For a basis $v_1, \dots, v_n$ of $V$, we have $\det f\not = 0$ iff $f(v_1\wedge \cdots \wedge v_n) = f(v_1) \wedge \cdots \wedge f(v_n) \not = 0$; that is, iff the $f(v_i)$ are linearly independent. Furthermore, since $h = fg$ has $\bar{h} = \bar{f}\bar{g}$, we have $\det(fg) = \det f \det g$. The other properties follow similarly. It requires a bit more sophistication than is usually assumed in a linear algebra class, but it's the first construction of $\det$ I've seen that's motivated and transparently explains what's otherwise a list of arbitrary properties.

The way I teach determinants to my students is to start with the case $n=2$, and to use the complex numbers and/or trigonometry in order to show that, for $(a,b), (c,d)$ vectors on the plane, the quantity $$ad-bc=||(a,b)||\cdotp ||(c,d)|| \sin \theta$$ is the signed area between $(a,b)$ and $(c,d)$ (in this order).

Then, using the vector product and its properties (we have seen it before coming to the topic of determinants in full generality), we check that $3$ by $3$ determinants carry the meaning of signed volumes.

The next step is to introduce determinants as alternate multilinear functions. We have seen examples of bilinear maps (inner products), trilinear maps, such as $$(u,v,w)\mapsto (u\times v)\bullet w$$ and the quadrilinear maps $$(a,b,c,d)\mapsto (a\bullet c) (b\bullet d)-(b\bullet c) (a\bullet d),$$ $$(a,b,c,d)\mapsto (a \times b)\bullet (c\times d).$$

Now, when explaining multilinearity we did emphasise that the fact that the last two examples are equal can be proven if only we check equality for the case where $a,b,c,d$ are vectors of the canonical basis.

Then the time comes to define the determinant of $n$ vectors in $\mathbb{R}^n$, which is a new example of $n-$linear, alternate function. They check that the vector space of such maps is indeed $\left(^n_n\right).$ The students thus learn that the determinant is essentially the only possible such function, up to a multiple, in the same way they saw that more general multilinear maps depend exclusively on their values on vectors of a chosen basis (say, the canonical basis in our case).

Although I learnt to prove stuff such as $\det(AB)=\det(A) \det(B)$ by strict functoriality, in class we do define the map $$L(X_1, \ldots , X_n)=\det(AX_1, \ldots , AX_n),$$ which by uniqueness is a constant multiple of the determinant function $T(X_1, \ldots , X_n)=\det(X_1, \ldots, X_n),$ and compute the constant by evaluating on the identity matrix, i.e. $X_i=e_i.$

Thus $\det(AB)=\det(A)\det(B).$

It is in T.W. Korner's book called Vectors, Pure and Applied that one can see a construction that uses elementary matrices and is rigorous. The OP can check Korner's book to see a nice, slightly more down-to-earth exposition.

In op. cit. one can see how Korner uses the fact that an invertible matrix can be decomposed as a product of elementary matrices to obtain the formula $\det(AB)=\det(A)\det(B).$

Note: I have been deliberately brief in my exposition, just so as not to repeat too much stuff that was already included in other answers.

$\det$ is the only multilinear alternating linear map such that $\det I = 1$. (2) and (3) combined with the following property would define $\det$ uniquely. $$ \det(a_1,\dots,u,\dots,a_n) + \det(a_1,\dots,\lambda v,\dots,a_n) = \det(a_1,\dots,u + \lambda v,\dots,a_n) $$ The only thing missing from your definition is linearity.