Here are some comments on the questions of the posting.
Every complex algebraic variety has the homotopy type of a finite CW-complex, so the Betti numbers are finite and the Euler characteristic is well defined. To simplify consider the quasi-projective case. Let $X\subset \mathbf{P}^n(\mathbf{C})$ be a projective variety and let $Y$ be a closed subvariety, then we can consider both $X$ and $Y$ as real algebraic varieties in $\mathbf{P}^{2n}(\mathbf{R})$. Now real projective spaces can be embedded in affine spaces, as opposed to the complex case: we can associate to a line $l$ in $\mathbf{R}^{k+1}$ the unique orthogonal projection with image $l$. A similar trick exists over an arbitrary field (the Jouanolou trick), but it is the uniqueness of the projection that gives us an embedding in the real case. In coordinates we take a point $(x_0:\cdots: x_k)$ to the $k+1$ by $k+1$ matrix $(\frac{x_i x_j}{\sum x_i^2})$.
This implies that real projective varieties are affine. Then one can use the triangulation theorem for real affine varieties, as presented e.g. in Hironaka's Arcata 1974 lectures to triangulate $X$ so that $Y$ is a subcomplex; this gives a triangulation of $X\setminus Y$ (infinite if $Y$ is nonempty) and a finite CW complex homotopy equivalent to $X\setminus Y$.
I'm pretty sure a similar result should hold for arbitrary (not necessarily projective) complex algebraic varieties and also for algebraic spaces, but I've never seen the details worked out in the literature.
The other question (whether or not the algebraic structure determines the cohomology) is not completely trivial either (and the answer depends on what exactly one means by cohomology). If $X$ is defined over a finite extension $F$ of $\mathbf{Q}$ and $X'=X\times_F\mathbf{C}$ for some embedding $F\subset\mathbf{C}$, then the cohomology ring of $X'(\mathbf{C})$ with finite coefficients does not depend on the embedding (and hence neither does the complex cohomology ring), see Freitag-Kiehl, \'Etale cohomology, theprem 11.6. It was an old question of Grothendieck whether the rational cohomology ring can depend on the embedding. It turned out even the real cohomology ring can, as recently shown by F. Charles. See www.math.ens.fr/~charles/crll5855.pdf
upd: here is an explicit procedure to obtain the Euler characteristic from the algebraic data: as mentioned in the comments, if $X$ is smooth and complete, we can take the alternating sum of the Euler characteristics of $\underline{\Omega^i}_X$'s. If $X$ the complement of a simple normal crossing divisor $D_1\cup\cdots \cup D_k$ in a smooth complete $D_{\varnothing}$, then $$\chi(X)=\sum_{I\subset\{1,\ldots,k\}}(-1)^{|I|}\chi(\cap_{i\in I} D_i).$$ Here we use the fact that the differentials in a spectral sequence do not change the Euler characteristic. In general one stratifies $X$ so that the difference of any two consecutive strata is smooth and takes the sum of the Euler characteristics of the strata. Using a similar procedure one can compute the Serre characteristic, which is a 2 variable analog; it can be seen as the image of the compactly supported cohomology in the Grothendieck group not of $\mathbf{Q}$-vector spaces, but of the mixed Hodge structures.
You can use étale cohomology to get a purely algebraic definition. (You can also use algebraic de Rham cohomology as Kevin suggests which has an extension to the singular non-proper case).
On way of seeing that the Euler characteristic is well-defined is to use the fact that $X$ is homotopy equivalent to a finite complex. (You can also use the finiteness theorem for cohomology of constructible sheaves.)