What's wrong with this code..?
I can't add comment to your question so I post as an answer even though this is not much an answer. This is just a hint to solve your problem.
8x8 led matrix and 595s are pretty low-level stuff, so I assume you have some knowledge about electronics and programming in general.
I think your wiring is OK but I also think you misunderstand how 595 works.
Here is the flow taken from a library (ESP32) for 64x64 RBG LED panel using bunch of 595s:
uint8_t data[8]; // data for one row

digitalWrite(OE_PIN, HIGH); // disable output, 595 outputs go into high impedance state

for (int i=0; i&lt;8; i++) 
  shiftOut(DATA_PIN, CLOCK_PIN, MSBFIRST, data[i]);  // dump data out

digitalWrite(LATCH_PIN, LOW);  // pulse 1 clock
delayMicroseconds(1); // note 2*
digitalWrite(LATCH_PIN, HIGH); // latch is active high 

digitalWrite(OE_PIN, LOW); // Enable output 

I think now you know how to fix your code.
Note 1: I find those names SER, SRCLK, RCLK (on TI's datasheet) are confusing. I prefer the names SERIAL DATA INPUT, SHIFT CLOCK, LATCH CLOCK (on ONSEMI MC74HC595 datasheet) or better DATA, CLK, LATCH, OE, RESET.
Note 2: Check 74HC595 datasheet to see clock pulse width limit and speed limit. And if you are using UNO or MEGA (@16MHz), you are good without stretching the clock (data clock and latch clock). If you use higher speed MCU, e.g: ESP32@240MHz or STM32@72MHz, you have to consider clock stretching there and also inside shiftOut() function.
To improve your circuit, you should wire both OE of 595s to a Digital pin. You can also connect both latch pin into one digital pin. Every time you latch, botch 595 latch, and every time you enable output, both enable output.
So with your configuration, just shiftOut data normally for the 595 on the left and invert data for the 595 on the right, like this
#define DATA_COL_PIN 11
#define DATA_ROW_PIN 5
#define CLOCK_COL_PIN 13
#define CLOCK_ROW_PIN 7

#define OE_PIN 4 //use digital pin 4 to enable output on both 595s
#define LATCH_PIN 6 // or 12, but you must connect both latch inputs of both 595s together

uint8_t col = 0b00000001; // for the left 595
uint8_t row = 0b00000001; // for the right 595

digitalWrite(OE_PIN, HIGH);
shiftOut(DATA_COL_PIN, CLOCK_COL_PIN, MSBFIRST, col);
shiftOut(DATA_ROW_PIN, CLOCK_ROW_PIN, MSBFIRST, ~row); // ~row = 0b11111110 = invert_bits(row)
digitalWrite(LATCH_PIN, LOW);
digitalWrite(LATCH_PIN, HIGH);
digitalWrite(OE_PIN, LOW);

The code above is just for you to understand how it works. Normally, they would work on a buffer such as
uint8_t row[8]; // each byte is one row

And they do row scanning, turn one row ON at a time and just flip each column ON and OFF following the byte data of that row.
I stop here as this is going off-topic. You just ask what is wrong with your code right?
BTW, 595 can sink/source 35mA on each pin (70mA in total). So if you want to light up one row at a time with your setup, you should choose each current limiter carefully so that not to exceed 70mA in total.

70mA / 8xLED = 8.75mA for each LED (assume red LED)
(5v - 1v7) / 8.75mA = 377R

It's likely the resistors should be at least 377 ohm or 470 ohm for safety.