Are SAT problems with at most two false clauses NP-complete?
Yes.  You can solve this in polynomial time.  The same strategy as in Are SAT problems with at most one false clause NP-complete? works.

Let $\varphi$ denote the CNF formula you care about, and $n$ the number of variables in $\varphi$.  Let $A_{i,j}$ denote the number of truth assignments that do not satisfy clauses $i$ or $j$ of $\varphi$, let $A_i$ denote the number of truth assignments that do not satisfy clause $i$ but do satisfy all other clauses of $\varphi$, and let $A$ denote the number of truth assignments that satisfy all clauses of $\varphi$.

These three cases are mutually exclusive.  Therefore,

$$\sum_{i,j} A_{i,j} + \sum_i A_i + A = 2^n,$$

where $n$ counts the number of variables in the formula.  In other words,

$$A = 2^n - \sum_{i,j} A_{i,j} - \sum_i A_i.$$

You can compute $A_i$.  Let $S_i$ denote the set of variables mentioned in clause $i$.  There is only one assignment to the variables in $S_i$ that will make clause $i$ false, so substitute in those values for the variables in $S_i$ and simplify the formula; the result is a CNF formula $\varphi'$ where we are guaranteed that every assignment to the remaining $n-|S_i|$ variables satisfies at most one clause, so https://cs.stackexchange.com/a/115849/755 tells us how to count the number of satisfying assignments to $\varphi'$, say $n_i$.  Then $A_i = n_i$.

You can compute $A_{i,j}$.  It is either $2^{n-|S_i \cup S_j|}$ or $0$, depending on whether there is any variable $x_k$ such that both $x_k$ and $\neg x_k$ appear in those clauses $i$ and $j$ of $\varphi$.

Therefore, you can compute $A$, the number of assignments that satisfy the formula.  Now, if $A \ne 0$, then you know that the formula is satisfiable; if $A=0$, the formula is not satisfiable.