In general, if $N$ is a finitely generated $S$-module, and $J$ is an ideal with $JN=N$, it is a standard fact that there exists $u\in J$ with $(1-u)N=0$. (This is one incarnation of Nakayama's Lemma.) In your context everything is finite-dimensional and therefore finitely generated. The condition $R/I\otimes_RI=0$ is equivalent to $I=I^2$, so there exists $e\in I$ with $(1-e)I=0$. In particular $(1-e)e=0$ so $e$ is idempotent. This means that there is a splitting $R=R_0\times R_1$ with $e=(0,1)$ and $I=0\times R_1$ so $R/I=R_0$. Now for any $R$-module $M$ the conditions $M\otimes_R(R/I)=0$ and $M\otimes_RI=M$ are equivalent and just mean that $M=0\times M_1$ for some $R_1$-module $M_1$. Your final condition $M\otimes_RM=M$ is then equivalent to $M_1\otimes_{R_1}M_1$. You can make this true by taking $M_1=R_1$, and I think that that is the only possibility. Of course you then have a split exact sequence $R/I\to M\to I$. So there are plenty of examples, but none of them are very interesting.

In general, if $N$ is a finitely generated $S$-module, and $J$ is an ideal with $JN=N$, it is a standard fact that there exists $u\in J$ with $(1-u)N=0$. (This is one incarnation of Nakayama's Lemma.) In your context everything is finite-dimensional and therefore finitely generated. The condition $R/I\otimes_RI=0$ is equivalent to $I=I^2$, so there exists $e\in I$ with $(1-e)I=0$. In particular $(1-e)e=0$ so $e$ is idempotent. This means that there is a splitting $R=R_0\times R_1$ with $e=(0,1)$ and $I=0\times R_1$ so $R/I=R_0$. Now for any $R$-module $M$ the conditions $M\otimes_R(R/I)=0$ and $M\otimes_RI=M$ are equivalent and just mean that $M=0\times M_1$ for some $R_1$-module $M_1$. Your final condition $M\otimes_RM=M$ is then equivalent to $M_1\otimes_{R_1}M_1$. You can make this true by taking $M_1=R_1$, and I think that that is the only possibility. But anyway, as $R/I=R_0\times 0$ and $M=0\times M_1$ we have $\text{Hom}_R(R/I,M)=0$, so there can be no short exact sequence $R/I\to M\to I$.