Idea is correct, proof lacks rigor, coefficient of $t$ in $\det(I+tA)$

The result is correct. I think your argument is fine, although it is not very amenable to be written nicely.

Here is a shorter proof.

If $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $A$ counting multiplicities, then the eigenvalues of $I+tA$ are $1+t\lambda_1,\ldots,1+t\lambda_n$. Thus $$ \det(I+tA)=(1+t\lambda_1)\cdots(1+t\lambda_n)=t^n\lambda_1\cdots\lambda_n+\cdots+t(\lambda_1+\cdots+\lambda_n)+1. $$ So the coefficient of $t$ is $$ \lambda_1+\cdots+\lambda_n=\operatorname{Tr}(A). $$ We also see from above that the coefficient of $t^n$ is $\det A$.

Here is a different argument. Let the columns of $A$ be $C_i$. Then the determinant viewed as a function of the sequence of columns is multilinear. Let $E_i$ be the columns of the identity matrix. Then you have $$\det(E_1+tC_1, \cdots , E_n+tC_n)$$ now by mulitlinearily this can be expanded like an binomial product, and thus the coefficient of $t$ is

$$\sum \det(E_1, \cdots , C_i,\cdots , E_n)$$ and it is easy to see that

$$ \det(E_1, \cdots , C_i,\cdots , E_n)=a_{ii}.$$