The choice of whether to require varieties in the scheme-theoretic sense to be (geometrically) irreducible and\or reduced seems to depend on the source. Certainly a $k$-variety ($k$ a field) should be of finite type and separated. The all-mighty Stacks Project also requires integrality (reduced+irreducible), but as Stacks observes, this has the disadvantage that the base change of a variety along a field extension need no longer be a variety (it can fail to be reduced or irreducible). Some sources (e.g. Milne's article in Cornell-Silverman on abelian varieties) require varieties to be geometrically integral. This ensures that base changes along field extensions and products of varieties are again varieties, defining away the problem to which Stacks alludes. I'd say it depends on the application you have in mind whether or not you require your variety to be (geometrically) irreducible. Reducedness I think is less controversial, although I believe Liu does not require his varieties to be reduced.
In some cases, certain things come automatically. For example, if $X$ is a $k$-scheme that is connected and has a $k$-point, or more generally a point $x$ such that $k$ is separably closed in $k(x)$, then $X$ is geometrically connected. So a connected $k$-group scheme (which has a $k$-point by definition, the identity section) is geometrically connected. If moreover the $k$-scheme is $k$-smooth (hence locally of finite type), then it will necessarily be geometrically integral ($X_{\overline{k}}$ is regular by the smoothness condition and this plus locally Noetherian plus connected implies irreducible). So, for example, in defining an abelian variety over $k$, which is usually defined as a proper smooth connected $k$-group scheme, you have something which is automatically geometrically integral.
Classically, I think varieties were pretty much always required to be irreducible (at least this is the case as you observe in Hartshorne). A variety which is integral has the property that the coordinate ring of any affine open is a domain, which is nice. But again, I don't think there is a conceptual reason when defining varieties over $k$ scheme-theoretically to insist that they be irreducible (and if you do, you might want to insist that they be geometrically so to ensure that products of varieties are varieties, etc.).
Also, on an unrelated note, when you write "...every projective variety in the sense of the above definition is of the type
$\textrm{Proj}\frac{k[T_1,\ldots,T_n]}{I}$
where $I$ is any homogeneous ideal," the word "any" should be replaced with "some" or "an." The way you have it worded makes it sound like any homogeneous $I$ cuts out the same closed subscheme of projective space.
Consider the integrally closed ring $R=S(Y)_{M_P}$, let $R_0$ be the set of its elements of degree $0$. Let $f=a/b \in Frac(R_0)$ be integral over $R_0$, thus $f$ is integral over $R$ so is in $R$. Thus, we have $a=fb$, $a,b \in R_0$, $f \in R$. As $R$ is integrally closed, $R$ is an integral domain, so $f$ cannot have any component of degree nonzero, thus $f \in R_0$.
Best Answer
This definition is fine and part (d) gives you a definition which is equivalent and doesn't encounter this issue of your. Your concern can be fixed with a little ingenuity.
Suppose $R=\bigoplus_d R_d$ is a graded domain, and let $K$ be its quotient field. Assume there exists a nonzero element $y\in R_1$. Let $K_d$ denote the subgroup of elements which can be represented as $\frac{r_1}{r_2}$ for $r_1,r_2$ homogeneous and $\deg r_1-\deg r_2=d$. Then $K'=\bigoplus_{d\in\Bbb Z} K_d$ is a subring of $K$, it's isomorphic to $K_0[y,1/y]$, and $K=K_0(y)=\operatorname{Frac} K'$. We observe that $K_0[y,1/y]$ is integrally closed in its field of fractions, being the localization of a polynomial ring over a field, so the integral closure of $R$ must be a subring of $K'$, and in particular consist of sums of homogeneous fractions. So we've eliminated your difficulty.