Let $A$ be an $n \times n$ real matrix. Suppose that each $e_i$ ($1 \leq i \leq n-1$) is in the direct sum of all eigenspaces of $A$ corresponding to eigenvalues with positive real part.
Here, $e_i$ denotes the vector with $i$th component $1$, and any other component $0$.
Now, suppose that I add some sufficiently large postive number to the $(n,n)$th entry $a_{nn}$ of $A$. Can I assure that the resulting matrix only has eigenvalues with positive real parts?
No. Consider $$ A=\pmatrix{-1&-2\\ 2&2}. $$ The two eigenvalues of $A$ are $\frac12(1\pm\sqrt{7}i)$. The direct sum of all eigenspaces of $A$ corresponding to eigenvalues with positive real parts is therefore $\mathbb C^2$, which certainly includes the standard basis as a subset. Thus all premises in your question are fulfilled. However, when $p>0$ is large, $\det(A+pE_{22})<0$ because $a_{11}<0$. Therefore $A+pE_{22}$ has a negative eigenvalue.
Edit. In general, denote by $B$ the leading principal $(n-1)\times(n-1)$ submatrix of $A$. Let $f$ and $g$ be the characteristic polynomials of $A$ and $B$ respectively. By expanding along the last row/column, we obtain $\det(xI-A-pE_{nn})=f(x)-pg(x)=p\left(\frac1pf(x)-g(x)\right)$. Therefore, when $p>0$ is getting larger and larger, some $n-1$ eigenvalues of $A+pE_{nn}$ will become arbitrarily close to the eigenvalues of $B$.
So, if all eigenvalues of $A+pE_{nn}$ have positive real parts when $p>0$ is large, all eigenvalues of $B$ must have nonnegative real parts. This gives us a necessary condition.
For sufficiency, suppose all eigenvalues of $B$ have positive real parts and $p>0$ is large. Then $\det(B)>0$ and $A+pE_{nn}$ has at least $n-1$ eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_{n-1}$ with positive real parts. Therefore $\det(A+pE_{nn})=p\det(B)+\det(A)>0$. It follows that the remaining eigenvalue $\lambda_n$ must have a positive real part too. For, if it is non-real, its complex conjugate must be some $\lambda_j\in\{\lambda_1,\ldots,\lambda_{n-1}\}$ and hence $\Re\lambda_n=\Re\lambda_j>0$. If $\lambda_n$ is real instead, since the other eigenvalues of $A+pE_{nn}$ must be real positive or appear in conjugate pairs, we have $\prod_{j=1}^{n-1}\lambda_j>0$. Since we also have $\prod_{j=1}^n\lambda_j=\det(A+pE_{nn})>0$, we conclude that $\lambda_n$ is a positive real number.